I have conjectured this result for the Frattini subgroup by doing some calculations in GAP. I think this is even true if $|G|=p_1^{i_1}\cdots p_n^{i_n}$ for $i_j\leq 3$ holds, but I would like to stick with this example first. I know the following:
If $p\mid |G|$, then $p\mid |G/\Phi(G)|$, thus the order of $\Phi(G)$ is at most $p^2q$. Since $\Phi(G)$ is nilpotent, it must be abelian, too.
That's why I think that it suffices to show this result for only abelian $G$ (which is fairly easy), as considering all abelian groups with that order would lead to all possible abelian $\Phi(G)$. But I still need a valid argument for that.
I also tried the assumption $\Phi(G)=C_p\times C_p$ or $\Phi(G)=C_p\times C_p\times C_q$ but I have not yet found a contradiction.
I would be glad to see some ideas or approaches for this conjecture.