Is the axiom of dependent choice appliable to a proper class? Let $X$ be a proper class and $a_0\in X$. Let $R$ be a binary relation on $X$. Then, does $\mathbf{ZF+DC}$ prove the existence of a sequence $\{a_n\}$ satisfying $a_n R a_{n+1}$? If not, does $\mathbf{ZF+AC}$ prove this?
Actually the above statement cannot be encoded in $\mathbf{ZFC}$ because it invokes a term “proper class”. Hence, let’s consider  a specific example.
Let $R$ be a ring. Assume the fact that “For every $R$-module $X$, there exists a free $R$-module $Y$ and an $R$-module epimorphism $\phi:Y\rightarrow X$”. Only assuming this, does $\mathbf{ZF+DC}$ prove the existence of a free resolution of a given $R$-module $M$?
 A: I think you want $R$ to be entire on $X$ (since otherwise take $R=\emptyset$). 
Assuming that, the answer to your general question is yes: let $\alpha$ be the least ordinal (or really any ordinal) such that $X\cap V_\alpha$ is infinite, and apply DC to $R\upharpoonright X\cap V_\alpha$.

A  key point here is that even though your instance $(X, R)$ is class-sized, the object we want to get out of it is set-sized. Whenever we have this situation, it's usually the case that the class version (appropriately phrased) follows from the set version by something like the argument above.
This isn't quite Scott's trick - Scott's trick is a method for representing equivalence classes which are possibly proper classes by sets in a canonical way - but it's related. 
A: That depends on what you mean by "class".
You could mean that it is any collection of sets, definable or otherwise. In that case, the answer is negative. For example, if $M$ is a countable model of $\sf ZFC$, then we can choose a cofinal $\omega$-sequence, $\alpha_n$, of ordinals in $M$ and define $R$ to be $x\mathrel{R}y$ if and only if for some $n$ the rank of $x$ is $\leq\alpha_n$ and the rank of $y$ is $>\alpha_n$.
If, however, you mean a definable class and $R$ is a total relation on $X$, then the answer is positive. The reason is that we can definably find the next rank of candidates for "continuing the sequence", and definably create a set of options for proceeding through $R$, and then $\sf DC$ is enough to prove this (note that in the requirements of $\sf DC$, the relation is total just as well).
To argue why, note that we define recursively the following classes: $X_0=\{a_0\}$; and $X_{n+1}$ is the collection of $x\in X$ with minimal rank such that there is some $y\in X_n$ such that $y\mathrel{R}x$.
The function $n\mapsto X_n$ is definable, and each $X_n$ is a set (by virtue of being bounded in rank); therefore the collection $\{X_n\mid n\in\omega\}$ is a set as well. So its union is a set, and we can now proceed with the usual $\sf DC$ arguments when applying this to the restriction of $R$ to the union $\bigcup_n X_n$.
