When $F$ is algebraically closed (or, more generally a splitting field), the answer is
positive in the sense that every irreducible right $FG$-module is isomorphic to a minimal
right ideal of the group algebra $FG$ (I stick to my preferred notation). Possibly the simplest argument I know, which really dates back to Richard Brauer is as follows:
(I assume the structure of semi-simple algebras known, which is reasonable for MO).
Let $V$ be an irreducible right $FG$ module, and let $\sigma:FG \to {\rm End}_{F}(V)$
be the associated representation. Let $\tau$ be the $F$-valued trace afforded by this representation. Form the element $i(\tau) = \sum_{g \in G} \tau(g^{-1})g$, which is
a central element of the group algebra $FG.$ Let $I_{\tau}$ denote the (two-sided) ideal
$i(\tau)FG$ of $FG.$ Notice that for each $h \in G$, we have
$i(\tau).h = \sum_{g \in G} \tau(g^{-1}h)g$. Hence for any $x \in FG$,
we have $i(\tau)x = \sum_{g \in G} \tau(g^{-1}x) g.$ In particular, $i(\tau)x = 0$
whenever $x \in J(FG)$, since nilpotent endomorphisms have trace 0. More generally,
the annihilator of $V$ (which is a maximal two-sided ideal of $FG$) annihilates $i(\tau)FG$.
Since $\{ g\sigma: g \in G \}$ spans ${\rm End}(V)$, we see that an element $x$ of $FG$
annihilates $V$ if and only if $\tau(g^{-1}x) = 0$ for each $g \in G$, so the annihilator
of $i(\tau)FG$ is no larger than the annihilator of $V$. Hence $i(\tau)FG$ is isomorphic
to the simple algebra ${\rm End}_{F}(V)$ just as as right $FG$-module, and the latter module is isomorphic to the direct sum of ${\rm dim}_F(V)$ copies of $V$ as right $FG$-module.
Hence a minimal right ideal of $FG$ contained in $i(\tau)FG$ is isomorphic to $V$.