Call $F : \mathbf C \to \mathbf D$ essentially injective (mirroring the definition of essentially surjective functor) if $F(A) \cong F(B)$ implies that $A \cong B$. This matches your definition: a functor between skeletal categories is essentially surjective if and only if its object function is injective.
Recall that a functor is conservative if, whenever $F(g) : F(A) \to F(B)$ is an isomorphism, $g : A \to B$ is also an isomorphism.
These two notions are related: in particular, a conservative functor that is full is also essentially injective, because if $F(g)$ is an isomorphism, then so is $g$ and hence $A \cong B$. Fullness ensures that every morphism between $F(A)$ and $F(B)$ is in the image of $F$, so if an isomorphism exists, it will necessarily be reflected. (As Mike Shulman points out in the comments, it suffices for $F$ to be pseudomonic, which means that it is faithful, and full on isomorphisms.)
Finally, every fully faithful functor is conservative, and hence essentially injective. It is easy to see that the converse does not hold: for instance, any functor from the category with two objects and a single isomorphism is essentially injective, but will rarely be fully faithful.