I had a similar question some time ago. See: Why need the morphisms to form a set ?

It turns out that there is no need to require that the morphisms of a category form a set. Therefore one can form the functor category $C^D$ for any categories $C, D$ in the usual way and it is again a category (those morhisms may form a proper class). The usual proof that $C^D$ is abelian if $C$ is carries over without difficulties.

BTW: If the objects of a category form a proper class and the morphisms form a set then the category is called "locally small".

One can use "big abelian categories". A "big abelian group" is a class that satisfies the same properties as an ordinary abelian group except that the underlying class doesn't need to be a set. Then a big abelian category is a category with the same properties as an ordinary abelian category except that the hom's aren't abelian groups but big abelian groups (see: Mitchell: "Theory of Categories", VII.1, page 164). 

The usual proof that $C^D$ is abelian if $C$ is abelian and $D$ is small carries over without difficulties to show that $C^D$ is (big) abelian if $C$ is abelian and $D$ is any category.