It is true for $g=1,2$. $Mod(S_1) \cong GL_2(\mathbb{Z})\cong Mod(S_{1,1})\cong Out(F_2)$.
In $Mod(S_2)$, the hyperelliptic involution is central and fixes the six Weierstrass points on any Riemann surface of genus 2. Since any mapping class group element commutes with this, it will permute the 6 Weierstrass points, and thus one may embed $Mod(S_2)\leq Mod(S_{6,2})$, the 6-punctured surface of genus 2 (this follows eg from Birman-Hilden theory). Hence $Mod(S_2) < Out(F_{9})$.
Notice that we are obtaining a section of the homomorphism
$Mod(S_{6,2})\to Mod(S_2)$. I wouldn’t expect such a section from $Mod(S_{k,g})\to Mod(S_g)$ for $g>3$. At least it’s not clear how one could do this using Weierstrass points, since the cardinality varies over moduli space.