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^*$$\widehat{\mathrm{Mod}(S)} \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$
has image Z^*$\mathbb{Z}^\star$, which is to say +-1$\pm1$. On the other hand,
Out(pihat) -> Sp_{2g}(Zhat) --det--> Zhat^*$$\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$
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)$\mathrm{Out}(\widehat \pi)$?"
By the way, is there a topological proof that Out(pihat) -> Zhat^*$\mathrm{Out}(\widehat \pi) \to \widehat{\mathbb{Z}}^\star$ is surjective? The only proof I know is that if you write down an algebraic curve X$X$ over Q$\mathbb{Q}$, the images of Frobenii in Out(pi_1^{et}(X_Qbar))$\mathrm{Out}(\pi_1^{et}(X_\overline{\mathbb{Q}}))$ 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)$\mathrm{Out}(\widehat \pi)$ whose determinant is not +-1$\pm1$!
