Yes, because $F$ has at most 2 sub-line bundles of degree $c$, so you have plenty of choices. The reason is the following. Twisting by a line bundle of degree $-c$ you reduce to the case $c=0$. You can assume that $F$ contains at least one sub-line bundle of degree $0$, and twisting again that this is $\mathcal{O}_C$, so that you have an exact sequence
$$0\rightarrow \mathcal{O}_C\rightarrow F\rightarrow L\rightarrow 0$$with $\deg(L)=1$. This extension is given by a nonzero extension class $e\in H^1(C,L^{-1})$.

Now suppose that $F$ contains another line bundle of degree $0$. It must map non-trivially to $L$, hence it is of the form $L(-q)$ for some point $q\in C$. This means that your extension splits when pulled back to $L(-q)$, or equivalently that the extension class $e$ goes to $0$ in $H^1(C,L^{-1}(q))$. Dually, $e$ defines a hyperplane $e^*$ in $H^0(C,K_C\otimes L)$, and this hyperplane must be equal to the image of $H^0(C,K_C\otimes L(-q))$; in other words, the image of $q$ by the map $\varphi :C\rightarrow |K\otimes L|^*$ defined by the linear system $|K\otimes L|$ must be equal to $e^*$. Since $\deg(K\otimes L)=2g-1$ it is an easy exercise to show that there are at most 2 such points $q$.

**Edit**: as explained in the comment below, there can actually be 3 such points if $C$ is hyperelliptic.