Hi there!
This is an interesting question.
The Riemannian mapping thm. actually gives a partial answer to that question - it is possible to find such a mapping, if the domain is not all of C, so the statement is seriously doubted - otherwise Riemann would have stated his thm in a more general way
However, I'll try to sketch a counter proof:
Assume that f: C -> D is holomorphic. It is evident that f is then an entire function. What is more, for all z in C it is true that |f(z)|<=1, i.e. f is bounded, because D is bounded. Now the Liouville's thm implies that f must be constant, implying the statement is incorrect.
(the Liouville's thm is that very nice tool also used for proving the Fundamental Theorem of Algebra in less that 5 lines)
regards,
marin