No. The fact that $\rho(c)$ has eigenvalues +1 and -1 implies that $\rho$ is defined over $L$. Here is a low-tech, longish argument (there are shorter ones using slightly more technology), which is essentially from an old paper of Wiles.

Work in a basis $(e_1,e_2)$ where $\rho(c)$ is the diagonal matrix $(1,-1)$. Now let $g \in G:= Gal( \overline {\bf Q}/\bf Q)$ and write $$\rho(g)=\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$$ in the same basis. Writing that $tr\ \rho(g)$ is in $L$ gives $a+d \in L$; writing that $tr \rho(gc)$ is in $L$ gives $a-d \in L$. Hence $a,d$ are in $L$ (for all $g \in G$).

Next takes a second element $g' \in G$, and write $$\rho(g')=\left(\begin{matrix} a' & b' \\ c' & d' \end{matrix}\right),$$
so that
$$\rho(gg') = \left(\begin{matrix} aa'+bc' & * \\ * & * \end{matrix}\right).$$
By the above paragraph applied to $g$, $g'$ and $gg'$, we know that $a$, $a'$ and $aa'+bc'$ are all in $L$. Thus $bc'$ is in $L$ (for all $g, g' \in G$, that is). Now we can certainly choose a $g$ such that $b \neq 0$ (otherwise $\rho$ would be reducible), and by changing the basis $(e_1,e_2)$ into $(be_1,e_2)$ (which does not affect the matrix of $\rho(c)$, hence does not affect the preceding conclusions), we may assume that $b=1$ for that $g$. We deduce that $c' \in L$ for all $g' \in G$. Next we can find a $g'\in G$ such that $c' \neq 0$, and then $bc' \in L$ implies $b \in L$. So we have shown that for all $g \in G$, all four coefficients of $\rho(g)$ are in $L$. QED.