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 injective 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, i.e. full on isomorphisms and faithful.)
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.