The question, as I said in comments to the answer by user72694, is very sensitive to the degree of smoothness. In Jim's question, the map $\varphi$ is only assumed to be $C^1$. I will work with maps of the unit square $I^2$ rather than the unit disk, but it is irrelevant since one can start with a square containing a disk in its interior. Recall that a $C^1$-map $f: I^2\to R^3$ is *short* if it does not increase length of tangent vectors. A map is *strictly short* if it strictly decreases the lengths of nonzero tangent vectors. It is very easy to construct strictly short maps, take, for instance, a map which is a dilation by a factor $<1$.

The key theorem then is:

**Theorem** (Nash-Kuiper). Every short $C^1$ embedding $f: I^2\to R^3$ can be approximated (in topology of uniform convergence) by ($C^1$) isometric embeddings.

See for instance this paper for a modern proof, using Gromov's technique of convex integration.

Now, alter the original map $f$ on a sub-square $Q \subset Int(I^2)$ a little bit and see that the approximation construction can be performed without changing the approximating maps away from an open neighborhood of $Q$. (If $f$ was a strictly short map, then altering $f$ on $Q$ with another strictly short map is easy since strictly short maps form an open set in $C^1$-topology.)

The drawback of this answer is that to verify that it works you would have to go through a proof of the N-K theorem, but if you are interested in $C^1$-isometric embeddings, you would have to read a proof anyway.