I'm still somewhat skeptical that this claim is true, do you have a good reason for believing it? If it is true here's a sketch of how the argument probably begins.
For p-groups, by the Burnside basis theorem a set generates G if and only if it generates $G/\Phi(G)$ where $\Phi(G)$ is the Frattini subgroup generated by commutators and pth powers (so $G/\Phi(G)$ is the universal elementary abelianization). This means that the rank is k if and only if the index of the Frattini subgroup is $p^k$.
Since $G/\Phi(G)$ is abelian, any subgroup containing $\Phi(G)$ is normal. This gives a large number of normal subgroups of $G_2$. So the idea would be to get an effective bound on the number of normal subgroups in $\Phi(G_1)$. But I don't see how to do this.
If such a theorem is true it's almost certainly somewhere in Huppert's Endliche Gruppen.