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".