If such an $L$-function with gamma factors whose constants have negative real parts exists, then it will have at most finitely many zeros inside the critical region because of the poles of these gamma factors. And in order to be symmetric about the real axis by virtue of the functional equation, these gamma factors must have either a purely real constant or come in pairs with conjugate constants. See e.g. the discussion after Proposition 1.9.1 in Bump's book Automorphic forms and representations (MR1431508, Zbl 0868.11022).
The way I see it, these will still be trivial zeros (i.e. they can be spotted by inspection of the arguments of the gamma factors), despite being located inside the critical region.
Note that the Euler product and the non-vanishing of the $L$-function on $\Re(s)=1$ prevents the trivial zeros from being symmetric outside the critical region and on its boundaries. For instance, Dirichlet $L$-functions of even characters have a trivial zero on $s=0$, with the notable exception of the RZF whose simple pole on $s=1$ cancels that of the gamma function on $s=0$, thus removing the singularity there, which is the reason why $\zeta(0)=-1/2$.
