This is a storm in a teacup. It is fueled by the conflation of two traces, which the experienced consider as self-explanatory, but confuses the novices. The exponent on your axial rotation on fermions lives in $$ \gamma_5 \otimes \vec \sigma = \begin{pmatrix} 1& 0 \\ 0 &-1\end{pmatrix} \otimes \vec \sigma, $$ in the Weyl basis, so it rotates R and L fermions in opposite directions in SU(2).
By contrast, the pion chiral field is just an SU(2) group element, $$ U = \exp\left(\frac{i}{F}\vec{\pi}\cdot{\sigma}\right). $$
You start with a trace summing over your large chiral component space and SU(2), for which I use the term Tr, but after you do the R/L components sum, you are left with a plain SU(2) matrix sum, for which I use tr, \begin{equation} \begin{split} \bar{Q}MQ &= -v^3\text{Tr} \left[ e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} M e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} \right]\\[0.25cm] &= -v^3 \text{Tr}\left[ e^{-\frac{i}{F}\gamma_5\vec{\pi}\cdot\vec{\sigma} }M\right]\\[0.25cm] &= -v^3\text{tr}\left [ (e^{-\frac{i}{F} \vec{\pi}\cdot\vec{\sigma}}+e^{\frac{i}{F}\gamma_5\vec{\pi}\cdot\vec{\sigma} } )M\right] =-v^3 \text{tr}\left [(U^\dagger +U)M \right ], \end{split} \end{equation}\begin{equation} \begin{split} \bar{Q}MQ &= -v^3\text{Tr} \left[ e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} M e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} \right]\\[0.25cm] &= -v^3 \text{Tr}\left[ e^{-\frac{i}{F}\gamma_5\vec{\pi}\cdot\vec{\sigma} }M\right]\\[0.25cm] &= -v^3\text{tr}\left [ (e^{-\frac{i}{F} \vec{\pi}\cdot\vec{\sigma}}+e^{\frac{i}{F} \vec{\pi}\cdot\vec{\sigma} } )M\right] =-v^3 \text{tr}\left [(U^\dagger +U)M \right ], \end{split} \end{equation} ready for your Gell-Mann—Oakes—Renner application.
Note that, even if M were m times the identity matrix, so, preserving vector isospin, the axial violation would still force your trace to involve a pion bilinear (mass term) proportional to it! So your pion mass squared will be proportional to m.