$\newcommand{\Ord}{\operatorname{\mathbf{Ord}}}$Yes, this is true. First, let us prove this if $C$ is a subclass of the ordinals. By transfinite recursion, we can define a function $f:C\to \Ord$ such that for each $c\in C$, $f(c)$ is the least ordinal greater than $f(d)$ for all $d\in C$ such that $d<c$. The image of $f$ is a (not necessarily proper) initial segment of $\Ord$: that is, it is either an ordinal or it is all of $\Ord$. Since $f$ is injective (it is strictly order-preserving), if the image of $f$ were an ordinal then $C$ would be a set by Replacement (using the inverse of $f$). Thus the image of $f$ is all of $\Ord$. But now it is trivial to find a subset of $C$ of cardinality $\alpha$: just take $f^{-1}(\alpha)$ (which is a set by Replacement).

Now let $C$ be an arbitrary proper class. Let $D\subseteq \Ord$ be the class of all ranks of elements of $C$. If $D$ is bounded, then it is contained in some ordinal $\alpha$, which means $C$ is contained in $V_{\alpha}$ and hence is a set. So $D$ must be unbounded, and is thus a proper class. By the previous paragraph, for any cardinal $\alpha$, there exists $S\subset D$ of cardinality $\alpha$. Now use Choice to pick a single element of $C$ of rank $s$ for each $s\in S$. The set of all these elements is then a subset of $C$ of cardinality $\alpha$.

(To be clear, this is a proof you can give for any particular class $C$ defined by some formula in the language of set theory. Of course, ZFC cannot quantify over classes, and so cannot even state this "for all $C$..." at once.)