Given that $\Gamma(s)$ and $\Gamma(1-s)$ are complex conjugates when $\Re(s)=1/2$, it is not surprising that $\Gamma(s)+\theta\Gamma(1-s)$ has an infinitude of zeros on the line $\Re(s)=1/2$, as long as $|\theta|=1$. The monotonicity argument given in the first answer then shows that there are no other zeros with $0<\Re(s)<1$. With the possible exception when the imaginary part of $s$ is small, the zeros for two different $\theta$ should interlace (if $\theta$ goes around the unit circle once, a zero is carried to an adjacent zero).