It is consistent with $\sf ZFC$ that every two proper classes have a bijection between them. That is to say, if $A$ and $B$ are two proper classes then there is a class $C$ which is a class of ordered pairs which is a bijection from $A$ onto $B$.
For example, if we assume that $V=L$, then this is true.
But it is also consistent that this is not the case, and for example there is no bijection between $\sf Ord$, the class of ordinals, and $V$, the class of all sets.
For your motivation, do note that in $\sf ZFC$, or even in class-able set theories like $\sf NBG$ or $\sf MK$ which extend $\sf ZFC$, one cannot talk about classes of proper classes.
We do something similar to How can one prove the axiom of collection in ZFC without using the axiom of foundation?, starting with a model of $\sf MK$ without Foundation and with a class of Quine atoms. We may assume that Global Choice holds in this model.
Next build a permutation model where the class of atoms is "set-amorphous" in the sense that every class of atoms is a set or its complement is a set. Easily, Global Choice fails.
Now suppose that you have a relation $R$ whose domain is a set, by the definition of the permutation model, it is fixed when we fix a set of atoms. Choose a large enough set of atoms $a$ fixing $R$, and now for every $x$ in the domain of $R$, if $y$ satisfies $R(x,y)$, then either $y$ is fixed by all the permutations fixing $a$, or its orbits under these permutations also satisfy $R(x,\pi y)$.
But now for every $x$ we can find a slightly larger $a_x$ which fixes also some $y$ such that $R(x,y)$. As the domain of $R$ is a set, there is a set $a$, without loss of generality, might as well assume that it was the firstly chosen set of atoms, containing all of these $a_x$'s, and thus we have that in this $V(a)$ we have enough representatives for choosing from $R$'s classes.
Finally, we can use the fact that $V(a)$ does satisfy a weak form of Foundation (that is by starting the von Neumann hierarchy with $a$ rather than $\varnothing$ we can generate $V(a)$) to ensure there is a Collection of images, and thus choice for sets is enough to finish the proof.
Note that this is an argument to show that Collection is weaker than Global Choice, as one would expect.
Best Answer
There are two options:
There is some $\alpha$ such that every $x\in X$ is equivalent to some $r\in V_\alpha$. In that case we normally treat $R$ as somehow having "set many equivalence classes" (e.g. algebraic field extensions or metric completions have these properties). Then you need only the axiom of choice for sets.
There is no such $\alpha$. In that case, we use the same trick, Scott's trick. Define $[x]_\sim$ to be the elements of $C$ equivalent to $x$ which have the least possible rank. Namely, let $\alpha$ be the least such that for some $r\in V_\alpha$, $x\sim r$, and $[x]_\sim=\{r\in V_\alpha\mid x\sim r\}$.
Now $\{[x]_\sim\mid x\in C\}$ is a class of sets, and the axiom of global choice can be applied.