Asymptotics expressions for $\Gamma$ were proposed in this thread :
\begin{align}
|\Gamma(x+iy)|&\sim \sqrt{2\pi}\,\exp\left[\frac{\left(x-\frac 12\right)\;\ln\bigl(x^2+y^2\bigr)}2-y\;\arg(x+iy)-x+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\
(1)\qquad&\sim \sqrt{2\pi}\,\bigl(x^2+y^2\bigr)^{\frac x2-\frac 14}\exp\left[-x-y\,\arg(x+iy)+\frac 1{12}\frac x{x^2+y^2}-\frac 1{360}\frac{x^3-3xy^2}{(x^2+y^2)^3}+\cdots\right]\\
\end{align}
(see too the end of this more detailed exposition)
Replacing your $\,q+1/2\,$ by $x$ and your $x$ by $y$ you want the asymptotics of :
$$\tag{2}\left|\frac{\Gamma(x + iy)}{\Gamma(1-x)}\right|^2=\left|\frac{\sin(\pi x)}{\pi}\right|^2\left|\Gamma(x)\,\Gamma(x + iy)\right|^2$$
where I used the reflection formula $(6.1.17)$ from A&S : $\;\displaystyle\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}\,$ .
Squaring $(1)$ twice (setting $y=0\,$ for $\Gamma(x)$ or using Stirling's formula) should give you the wished asymptotic expressions.
Visual representation of the logarithm of your function :
