The simplest case, which is already quite complicated, is if $F$ is algebraically closed of characteristic $0$ and $G$ is finite, or equivalently if $R$ is finite dimensional over $F$ (note that $R$ is necessarily an $F$-algebra). Then $F[G]$ is semisimple by Maschke's theorem, and furthermore splits up as a finite product of matrix algebras $M_n(F)$, one for every irreducible representation of $G$ over $F$ of dimension $n$. So in this special case the question reduces to asking:
Which tuples $(n_1, \dots n_k)$ arise as the dimensions of the irreducible representations of a finite group $G$ over $F$?
I don't think there's any hope of a simple answer to this question. Here are some necessary conditions. Note that $|G| = \sum n_i^2$.
- At least one of the $n_i$ must be equal to $1$ (since the trivial representation is always irreducible), and in fact the number of $n_i$ equal to $1$ must divide $|G|$ (since it's the order of the abelianization).
- If $|G|$ is prime then each of the $n_i$ must be equal to $1$ (since in this case $G$ is a cyclic group of prime order).
- Each of the $n_i$ must divide $|G|$.
For example, $F \times M_2(F)$ is not a group algebra (because $1^2 + 2^2 = 5$ and the only group of order $5$ is $C_5$), but $F \times F \times M_2(F)$ is (it's the group algebra of $S_3 \cong D_3$).