Here is a proof based on the idea of Richard Lyons.

**Proposition** Let $A$ be a finite abelian $p$-group on which a group $X$ acts. 
  Suppose that that for every $X$-invariant subquotient $B/C$ of $A$ $($i.e., both $B$ and $C$ are $X$-invariant) 
  such that $B/C$ is elementary abelian, $X$ acts completely reducibly on $B/C$. 
  Then $A$ is a direct product of $X$-invariant homocyclic subgroups.

**Proof**  Let $A$ be a counterexample of minimal possible order. 
  Then $A$ is $X$-indecomposable.
  Let $\overline{A}=A/\Phi(A)$, and observe that $X$ acts completely reducibly on $\overline{A}$.
  Then 
  $$\overline{A}=\overline{B}\times \overline{C}$$
  where $\overline{B}$ and $\overline{C}$ is completely reducible such that $\Phi(A)=B\cap C$, $\exp(B)=\exp (A)$.
  By the minimality of $A$, $B$ and $C$ are both direct product of $X$-invariant homocyclic subgroups,
  and, without loss of generality, we may assume that $B$ is homocyclic.
  Also, $\Phi(B)=\Phi(A)$.
  Let $F$ be a maximal homocyclic subgroup of $A$ containing $B$.
   Then $\Phi(B)\leq \Phi(F)\leq \Phi(A)$.
  However, since $\Phi(B)=\Phi(A)$, $\Phi(F)=\Phi(B)$, and hence $B=F$ (as $F$ is homocyclic).
  By Krull-Remak-Schmidt's theorem, $A=B\times V$ where $V$ is elementary abelian (as $\Phi(A)=\Phi(B)$),
  i.e. $A=B\Omega_1(A)$, where $\Omega_1(A)$ is a subgroup of $A$ generating by every element of order 2 in $A$.
  By the assumption of this proposition, $\Omega_1(A)$ is completely reducible.
  Therefore, $\Omega_1(A)=\Omega_1(B)\times D$, where $\Omega_1(B)$ and $D$ are $X$-invariant.
  Observe that $B\cap D=\Omega_1(B)\cap D=1$, and hence $A=B\times D$ is $X$-decomposible, a contradiction.

I guess Dave's idea is trying to show that: under the assumption of this question, every $\mathbb{F}_p[FH]$-module is completely reducible.