**Short Mathematical Answer**

No. The situation is quite different.

Original Monty Hall problem: Assume you pick door A. Then Monty shows either door B or door C to have a goat. And by symmetry, the chances of that are the same, i.e.

$$P(RevealGoat_B)=P(RevealGoat_C)=\frac{1}{2}$$

In fact, the conditional probability of showing $C$ to be empty if the car is behind door $A$ is the same as well:

$$P(RevealGoat_C|Car_A)=\frac{1}{2}$$

Therefore, if we assume that Monty shows door C to be empty, the chance that the car is behind door A is:

$$P(Car_A|RevealGoat_C) = \frac{P(RevealGoat_C|Car_A)\cdot P(Car_A)}{P(RevealGoat_C)}=\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}$$

which is why you should switch to door $B$.

Your Situation:

Assume you pick taco $A$ and that your wife will reveal contents of taco $C$ no matter what. The chance that she ends up revealing beans in taco $C$ is therefore simply the a priori chance of there being beans in taco $C$, i.e.

$$P(RevealBeans_C) = \frac{2}{3}$$

while the conditional probability chance that she ends up revealing beans in taco $C$ if the meat is in taco $A$ is of course:

$$P(RevealBeans_C|Meat_A) = 1$$

And therefore, after she showed you that there were beans in burrito $C$, the chance that your taco $A$ has the meat is:

$$P(Meat_A|RevealBeans_C) = \frac{P(RevealBeans_C|Meat_A)\cdot P(Meat_A)}{P(RevealBeans_C)}=\frac{1\cdot \frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$$

So no, not the same situation, and switching tacos does not help.

**Long, more Detailed and more Conceptual Answer**

The key is to realize that it is not just knowing that $C$ contains the beans or, as in the Month Hall problem, that $C$ contains a goat, but *how you found out that information*.

In the original Monty Hal problem, you know that Monty Hall knows where the prize is, and that he will show a goat after you pick, and that, if you happen to initially pick the door with the prize, Monty will randomly pick between the two goats to show to you. Indeed, there is no a priori reason for Monty to open door $C$, as Monty could have opened up door $B$ as well (assuming there was no car behind door $B$). So it's with all of that dynamics of the situation, and your knowledge of that dynamics, that you can figure out that it is better to switch. Indeed, see the Variants section in the Wikipedia page on the Monty Hall problem for how different behaviors and intentions on Monty's side will change whether switching is a good idea or not,

None of the dynamics behind the original Monty Hall problem were in place in the situation where you found yourself with the tacos and your wife. Your wife only knew the contents of taco $C$, and was only able to reveal its contents (that is, your wife was not going to just open up taco $A$ or $B$, and even if she did, she might have opened up one with the meat, unlike Monty Hall, who always shows a goat). In fact, it wasn't even clear that your wife was definitely going to 'spill the beans' on $C$! (forgive my pun ...)

So no, this is definitely not like the Monty Hall scenario.

Mathematically, the difference with the Monty Hall problem is as follows:

In the original Monty Hall, we have that:

$C_A$: event that $A$ (the one you picked) has the car

$C_B$: event that $B$ has the car

$C_C$: event that $C$ has the car

$ShG_C$: event that Monty shows $C$ to have a goat

$P(ShG_C|C_A) = \frac{1}{2}$ (if $A$ has the car, Monty will randomly pick between $B$ and $C$ to show a goat)

$P(ShG_C|C_B) = 1$ (if $B$ has car, then Monty definitely shows $C$, given that you picked $A$)

$P(ShG_C|C_C) = 0$ (of course, it is impossible for Monty to show a goat in $C$ if the car is in $C$!)

Hence:

$$P(ShG_C)=$$

$$P(ShG_C|C_A)\cdot P(C_A)+ P(ShG_C|C_B)\cdot P(C_B)+P(ShG_C|C_C)\cdot P(C_C)=$$

$$\frac{1}{2}\cdot \frac{1}{3} + 1\cdot \frac{1}{3} + 0\cdot \frac{1}{3} =\frac{1}{2}$$

(of course! Monty is always going to show one of the other two doors, and given the symmetry between $B$ and $C$, the probability of showing $C$ is $\frac{1}{2}$)

And therefore:

$$P(C_A|ShG_C)=\frac{P(ShG_C|C_A) \cdot P(C_A)}{P(ShG_C)}=\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}$$

i.e. the chance that the one you picked has the car is $1$ out of $3$, and hence you should switch!

But in this scenario, given that all your wife knows is the content of $C$, we have:

$M_A$: event that $A$ (the one you picked) has the meat

$M_B$: event that $B$ has the meat

$M_C$: event that $C$ has the meat

$ShB_C$: event that your wife shows $C$ to have beans

Now, what is the chance that your wife was going to show that $C$ has the beans? We don't really know of course, because this depends on her intentions. Maybe your wife was only going to tell you that $C$ has the beans if it contained the beans, and if $C$ contained the meat, she might have said nothing at all, just to keep you guessing some more. In fact, even if $C$ contained the beans, she may have chosen not to say anything. Then again, she might have told you what's in $C$ regardless of the contents of $C$: "Aww, you picked $A$; that's too bad, since $C$ contains the meat! You lose!"

In other words, in contrast to the Monty Hall scenario, we don't really know $P(ShB_C|M_A)$ or $P(ShB_B|M_B)$ (though like the Monty Hall scenario, we do have that $P(ShB_B|M_C)=0$ of course: it is impossible for your wife to reveal $C$ has beans if it has the meat!).

However, given that your wife knows nothing about the contents of $A$ and $B$, we can assume that

$$P(ShB_C|M_A)=P(ShB_B|M_B)$$

This is the crucial difference with the Monty Hall scenario, as in the Monty Hall scenario we have

$$P(ShG_C|C_A) \not = P(ShG_C|C_B)$$

OK, but do we know if switching was going to help or hurt you? Or, given that you don't know what your wife is going to do, maybe we can't tell?

Well, let's define:

$B_C$: event that $C$ has beans

In your taco scenario, we can say that:

$$P(ShB_C|M_A)=P(ShB_C|M_B)=P(ShB_C|B_C)$$

for in all three cases, your wife would be looking at beans in $C$, and decides to reveal that with probability $P(ShB_C|B_C)$.

So, we now have that:

$$P(ShB_C)=$$

$$P(ShB_C|M_A)\cdot P(M_A)+ P(ShB_C|M_B)\cdot P(M_B)+P(ShB_C|M_C)\cdot P(M_C)=$$

$$P(ShB_C|B_C)\cdot \frac{1}{3} + P(ShB_C|B_C)\cdot \frac{1}{3} + 0\cdot \frac{1}{3} =P(ShB_C|B_C)\cdot \frac{2}{3}$$

Note that in the case where your wife was definitely going to reveal that $C$ has the beans if $C$ has the beans, we have $P(ShB_C|B_C)=1$, and thus

$$P(ShB_C)=\frac{2}{3}$$

again in contrast to the Monty Hall scenario, where we have that

$$P(ShG_C)=\frac{1}{2}$$

Now, from your later edit, it seems like your wife would indeed always reveal the beans when she was looking at the beans, since she wanted you to find the one with the meat, but in fact you don't have to know that, because:

$$P(M_A|ShB_C)=\frac{P(ShB_C|M_A) \cdot P(M_A)}{P(ShB_C)}=\frac{P(ShB_C|B_C) \cdot \frac{1}{3}}{P(ShB_C|B_C)\cdot \frac{2}{3}}=\frac{1}{2}$$

And so, there you have it: in your situation, switching would not have made any difference.

.. and of course it shouldn't have! With revealing the contents of taco $C$ being the only possible move for your wife (as opposed to opening up taco $A$ or $B$), your wife was simply cancelling one of the options, leaving two with equal probability. And indeed, the same would be true if in the Monty Hall scenario Monty Hall would be restricted to revealing what's behind door $C$ from the start as well. Yes, in that scenario you would be helped by knowing that a goat is behind door $C$ (and hence you can increase your chances from $\frac{1}{3}$ to $\frac{1}{2}$) but if Monty had the freedom to reveal a different door, you would be helped even more. And to see that, consider what would have happened if door $C$ would have the car behind it, but Monty was restricted to only being able to reveal what's behind door $C$. Well, then Monty would have revealed the car, and you would have lost. But by having the freedom to open either door, Monty was sure to be able to pick one that is empty, thereby giving you 'maximal' additional information, and thus allowing you to increase your chances from $\frac{1}{3}$ to $\frac{2}{3}$ by switching to the other door.