This is surely not the most direct answer. But $\mathrm{Out}(\Pi_{g,1}) \cong \mathrm{Out}(F_{2g})$ surjects onto $\mathrm{GL}(2g,\mathbf Z)$, and the image of $\Gamma_{g,1}$ lands in $\mathrm{Sp}(2g,\mathbf Z)$. So the index is infinite for $g \geq 2$.

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In fact a more careful version of this argument shows that $(g,n)=(1,1)$ is the unique case where you get a finite index subgroup. Let me flesh out the argument. For all $n>0$ we have similarly that $\mathrm{Out}(\Pi_{g,n}) \cong \mathrm{Out}(F_{2g+n-1})$ surjects to $\mathrm{GL}(2g+n-1,\mathbf Z)$. The corresponding representation $H$ of $\Gamma_{g,n}$ is just the action of the mapping class group on the first homology of your favorite genus $g$ surface $\Sigma$ with $n$ punctures. But the action must be compatible with a lot of extra structure coming from geometry: there is the short exact sequence
$$ 0 \to \mathbf Z^{n-1} \to H_1(\Sigma,\mathbf Z) \to H_1(\overline \Sigma,\mathbf Z) \to 0 $$
coming from the long exact sequence of a pair, and $\Gamma_{g,n}$ preserves it, where $\overline \Sigma$ is the compact surface obtained by filling in the punctures. Moreover, $H_1(\overline \Sigma,\mathbf Z)$ is of rank $2g$ and carries a symplectic form preserved by $\Gamma_{g,n}$; the action of $\Gamma_{g,n}$ on $\mathbf Z^{n-1}$ is trivial. But $\Gamma_{g,n}$ having finite index in $\mathrm{Out}(\Pi_{g,n})$ would imply that the image of the mapping class group has Zariski closure all of $\mathrm{GL}(2g+n-1)$ or $\mathrm{SL}(2g+n-1)$. This contradicts the above unless $n=1$ (so the first term vanishes) and $g=1$ (so $\mathrm{Sp}(2g)=\mathrm{SL}(2g)$).