This is one of my old unsolved questions when I reading Novikov's book on homology theory. I do not know how to prove it because standard triangulation, fundamental diagram, etc does not help and it should be easy to prove.
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This question just floated up again, so let me put some references here.
This MO question gives several different proofs that for your surface $S$, $\pi_1(S)$ is free. In fact, Lee Mosher's answer gives a direct proof that $S$ is homotopy equivalent to a graph, and hence to a bouquet of circles. You can also proceed by noting the universal cover of $S$ is contractible, and hence $S$ is homotopy equivlanent to any $K(\pi_1(S),1)$, of which the appropriate bouquet of circles is one.
I would agree with Leonid. The claim should definitely be true for open surfaces which are built from closed ones by removing a top handle (i.e. a disc), (this should follow from a handle decomposition and the classification of closed surfaces). For example (as also stated in wiki if you search for surfaces) you could take a cantor set in the sphere, and take its complement. I don't expect this to have a good homotopy type.