Let $Z'$ be Zermelo set theory without Choice, and Separation replaced by Replacement.
Question 1: Is Separation a theorem of $Z'$ (since classical logic is assumed, LEM holds)?
Question 2: Is the following equivalence a theorem of $Z'$: Choice $\equiv$ 'Every entire relation from $\mathrm A$ to $\mathrm B$ (where $\mathrm A$,$\mathrm B$ are classes and every element of $\mathrm A$ is related to at least one element of $\mathrm B$) contains (the graph of) a function $\mathrm A$ $\rightarrow$ $\mathrm B$'? (Note: the form of this equivalence is taken directly from the nLab entry "Choice Object". Also, the Wikipedia article on Zermelo Set Theory suggests (in the subsection titled "The axiom of separation") that Zermelo's disposal of Russell's paradox in his system opens the door for talk of proper classes, and thus classes in general).
The answer to question (1) is: yes, as long as there is a way to produce the empty set without using replacement, e.g. via a special axiom. You can separate any non-empty subset by taking an element from it, and replacing every element you want to remove from the larger set with that specified element.
For question (2), I think you need to be more precise with what a relation of classes is -- the equivalence holds if you restrict $A,B$ to be sets. (For one direction, well-order $B$ and map an element of $a$ to the $B$-least element to which it is related. The other way around: using the condition, it is trivial to construct sections to all surjections.)