No. Consider the case of the trivial group scheme over a field $k$ (so $R=k$). In this situation, a quasi-coherent sheaf is just a $k$-vector space. As Lennart Meier said in a comment, you need additional structure to get an algebra, e.g., a multiplication map.

**Added:** If you just want a "pointwise" multiplication operation on your module $M$, you should ask for an $R$-linear map $M \otimes M \to M$. This does not use the group law $m: G \times G \to G$ (i.e., the coalgebra structure on $R$). It sounds like you might be looking for some kind of convolution product that uses the group law, for example, a map $F \boxtimes F \to m^*F$ on $G \times G$. Again, this is not a condition, but an extra structure. If we return to the case where $G$ is trivial, we see that the only conditions on a $k$-vector space that endow it with a canonical algebra structure are those conditions that imply the vector space is zero.