Hey Mark,
For $S_g$, I think you only need kernels of the first homology $\mod 2$ of all index $2$ subgroups of $\pi_1(S_g)$ (I haven't thought about the punctured case yet). Call this subgroup $K$. Suppose $h$ acts trivially on $\pi_1(S_g)/K$. First of all, notice that $h$ acts trivially on $H_1(S_g,\mathbb{Z}/2)$ (all further coefficients will be understood to be in $\mathbb{Z}/2$). If $\gamma$ is non-separating separating in $S_g$, then it divides $S_g$ into subsurfaces $A$ and $B$, say with $\chi(A)>\chi(B)$. Then since $h\gamma\cap \gamma=\emptyset$, we have $h(A)\cap A=\emptyset$, which means that $h(H_1(A))\neq H_1(A)$, a contradiction.
Now suppose that $\gamma$ is non-separating. Then since $h$ acts trivially on $H_1(S_g)$, $[\gamma]=[h(\gamma)]\in H_1(S_g)$. So $\gamma\cup h\gamma$ divides $S_g$ into subsurfaces $A$ and $B$. Consider the 2-fold cover $\tilde{S_g}\to S_g$ dual to $\gamma$, so that the preimage of $\gamma$ consists of two curves, $\tilde{\gamma}$ which separate $\tilde{S_g}$. Let $\tilde{h}$ be a the lift of $h$ to $\tilde{S_g}$. \tilde{S_g}$ which acts trivially on $H_1(\tilde{S_g})$ (which exists since by hypothesis $h$ acts trivially on $\pi_1(S_g)/K$). Then $\tilde{\gamma}\cup \tilde{h}\tilde{\gamma}$ divides $\tilde{S_g}$ into regions $\tilde{A_1},\tilde{A_2}$ and $\tilde{B_1},\tilde{B_2}$ which are each trivial covers of $A$ and $B$, respectively (labeled so that $\tilde{A_i}\cup \tilde{B_i}$ bounds $\tilde{\gamma}$, $i=1,2$). Then one sees that $\tilde{h}(\tilde{A_1}\cup\tilde{B_1})=\tilde{A_1}\cup \tilde{B_2}$, possibly up to relabeling. Then $\tilde{h}(H_1(\tilde{B_1}))=H_1(\tilde{B_2})$, \tilde{h}(H_1(\tilde{B_1}))\not=H_1(\tilde{B_1})$, which contradicts that $\tilde{h}$ acts trivially on $H_1(\tilde{S_g})$ (since it acts trivially on $\pi_1(S_g)/K$).
I'm not sure if an element of the Torelli group can have this property.
