The answer to the question as asked is definitely "no" in general. For any value of $n>1 $ and any odd prime $p$, we may take a perfect group $G$ which is a semidirect product of the form $E.{\rm Sp}(2n,p),$ where $E$ is extra special of order $p^{2n+1}$. Then $E$ is the solvable radical of $G$, and is non-Abelian, so is not (isomorphic to) a subgroup of any Schur multiplier.

The answer to the question as asked is definitely "no" in general. For any value of $n>1 $ and any odd prime $p$, we may take a perfect group $G$ which is a semidirect product of the form $E.{\rm Sp}(2n,p),$ where $E$ is extra special of order $p^{2n+1}$ and the action of the given symplectic group on $E$ is the natural one. Then $E$ is the solvable radical of $G$, and is non-Abelian, so is not (isomorphic to) a subgroup of any Schur multiplier.