First of all, this question came to me (or someone else with my initials) via Mark Kisin, so I can't claim credit (and for all I know it came to him from elsewhere.)

Second:  there's one obvious obstruction to surjectivity.  Namely, the map

hat{Mod(S)} -> Sp_{2g}(Zhat) --det--> Zhat^*

has image Z^*, which is to say +-1.  On the other hand,

Out(pihat) -> Sp_{2g}(Zhat) --det--> Zhat^*

is surjective.  So the map you ask about is definitely not surjective.  The question is whether in some sense "this is the only way the map fails to be surjective."  Since I don't have a precise meaning in mind for the phrase in quotes, one might just say "what is the closure of the image of the mapping class group in Out(pihat)?"

By the way, is there a topological proof that Out(pihat) -> Zhat^* is surjective?  The only proof I know is that if you write down an algebraic curve X over Q, the images of Frobenii in Out(pi_1^{et}(X_Qbar)) give you automorphisms of pihat with lots of different determinants.  Other than this I don't know how to construct a single element of Out(pihat) whose determinant is not +-1!