We prove that
$$I=-\frac{\pi}{4}(\gamma+\log 4).$$
$$I=\int_0^\infty\frac{t\arg\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt.$$
$I$ is the imaginary part of the complex integral
$$\int_0^\infty \frac{t\log\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt$$
using the usual branch of the logarithm of $\Gamma(s)$. Integrating by parts
$$=\frac12\int_0^\infty \log\Gamma(\tfrac14+\tfrac{it}{2})\,d\Bigl\{-\frac{1}{\frac14+t^2}\Bigr\}=
\Bigl.-\frac{1}{2(\frac14+t^2)}\log \Gamma(\tfrac14+\tfrac{it}{2})\Bigr|_{t=0}^\infty+\frac12\int_0^\infty \frac{1}{\frac14+t^2}\frac{i}{2}\frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\,dt$$
$$=2\log\Gamma(1/4)+\frac{i}{4}\int_0^\infty \frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\frac{dt}{\frac14+t^2}.$$
We have
$$-\frac{\Gamma'(s)}{\Gamma(s)}=\gamma+\frac{1}{s}+\sum_{n=1}^\infty\Bigl(\frac{1}{s+n}-\frac{1}{n}\Bigr).$$
It is easy to justify the interchange here so that
$$=2\log\Gamma(1/4)-\frac{i}{4}\Bigl\{\gamma\int_0^\infty \frac{dt}{\frac14+t^2}+
\int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}+\sum_{n=1}^\infty
\int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2}
\Bigr\}$$
With Mathematica we find
$$\int_0^\infty \frac{dt}{\frac14+t^2}=\pi,\quad \int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}=2\pi-4i,$$
$$\int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2}=-\frac{\pi+i\log(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-\frac14\Bigl\{\pi\gamma+2\pi-\sum_{n=1}^\infty \frac{\pi}{n(2n+1)}\Bigr\}$$
For the sum in $n$ we get with Mathematica
$$\sum_{n=1}^\infty\frac{1}{n(2n+1)}=2-2\log 2.$$
It follows that
$$I=-\frac14 \bigl\{\pi\gamma+2\pi-2\pi+2\pi\log2\bigr\}=-\frac{\pi}{4}(\gamma+\log 4).$$

The evaluations with Mathematica are not difficult to prove.