I had a similar question some time ago. See: [http://mathoverflow.net/questions/48810/why-need-the-morphisms-to-form-a-set][1]

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". 


  [1]: http://mathoverflow.net/questions/48810/why-need-the-morphisms-to-form-a-set