Yes. In fact, $V$ has an infinite dimensional semisimple quotient, which is decomposable since any simple $\mathbb{F}_pG$-module is a quotient of $\mathbb{F}_pG$ and so is finite-dimensional.
Let $V'=\operatorname{rad}(V)=V.\operatorname{rad}(\mathbb{F}_pG)$. Then $V/V'$ is infinite dimensional if $V$ is, and is semisimple. (It's infinite dimensional since any epimorphism from a finite dimensional projective module to $V/V'$ would lift to an epimorphism to $V$.)
For finitely generated profinite groups, as asked about in comments, there are counterexamples. I asked a former colleague, John MacQuarrie, who's worked on modular representations of profinite groups, and the following example is based on his answer (although any errors introduced in translating it into terms I understand are my own work).
Let $V$ be a vector space over $\mathbb{F}_p$ with countable basis $\{e_1,e_2,\dots\}$, and let $\mathbb{Z}$ act on $V$ by letting a generator send $e_i$ to $e_{i-1}+e_i$ for $i>1$ and $e_1$ to $e_1$. If $q$ is a power of $p$ with $q>i$, then $q$ fixes $e_i$, so the action extends to a discrete action of the $p$-adic integers $\mathbb{Z}_p$. It easy to check that the span $V_n=\langle e_1,e_2,\dots,e_n\rangle$ is a submodule for any $0\leq n$, and that these are the only proper submodules. This module can be more naturally described as the Pontryagin dual of the regular representation of the profinite group algebra $\mathbb{F}_p[[\mathbb{Z}_p]]$.