OK, here is (I hope) a proof. I prefer to consider the integral $\int_{\mathbb D}\left|\frac 1{z-a}+\frac 1{z-\bar a}\right|\,d\mu(z)$ where $a=e^{i\theta}, \theta\in(0,\pi/2)$.
This is just a one-parameter problem with explicit functions, so the first instinct is just to differentiate with respect to the parameter. However, if you try to directly differentiate with respect to $\theta$, you'll end up with a singular integral near the poles and all I can say after that is "Good luck with estimating anything". So we'll do it a bit more intelligently staying with absolutely convergent integrals all the way through. To this end, just move both poles and variable from a fixed position by the Mobius transformation $\zeta\mapsto \tau(\zeta)=\frac{\zeta-\varepsilon}{1-\varepsilon\zeta}$. The Jacobian is $J(z)=\frac{(1-\varepsilon^2)^2}{|1-\varepsilon z|^4}$ and we have
$$
\tau(z)-\tau(a)=\frac{(1-\varepsilon^2)(z-a)}{(1-\varepsilon z)(1-\varepsilon a)}\,,
$$
so
$$
\int_{\mathbb D}\left|\frac 1{\tau(z)-\tau(a)}+\frac 1{\tau(z)-\tau(\bar a)}\right|J(z)\,d\mu(z)\\
=
\int_{\mathbb D}\frac{1-\varepsilon^2}{|1-\varepsilon z|^3}\left|\frac {1-\varepsilon a}{z-a}+\frac {1-\varepsilon \bar a}{z-\bar a}\right|\,d\mu(z)
$$
We shall rewrite
$$
\frac {1-\varepsilon a}{z-a}+\frac {1-\varepsilon \bar a}{z-\bar a}
\\
=\frac{1+\varepsilon(z-a)}{z-a}+\frac {1+\varepsilon(z-\bar a)}{z-\bar a}-\varepsilon z\left(\frac 1{z-a}-\frac 1{z-\bar a}\right)
\\
=(1-\varepsilon z)\left[\frac 1{z-a}-\frac 1{z-\bar a}+\frac{2\varepsilon}{1-\varepsilon z}\right]
$$
resulting in the integral
$$
\int_{\mathbb D}\frac{1-\varepsilon^2}{|1-\varepsilon z|^2}\left|F(z)+\frac{2\varepsilon}{1-\varepsilon z}\right| d\mu(z)
$$
where $F(z)=\frac 1{z-a}+\frac 1{z-\bar a}$. Now we just differentiate with respect to $\varepsilon$ at $\varepsilon=0$ and see that all we need to prove is that
$$
\int_{\mathbb D}\left[|F(z)|\Re z+\frac{\Re F(z)}{|F(z)|}\right]\,d\mu(z)<0\,.
$$
Our next task will be to honestly
compute $\Re F(z)$. We have
$$
\Re F(z)=\Re\left[\frac{\bar z-\bar a}{|z-a|^2}+\frac{\bar z- a}{|z-\bar a|^2}\right]
\\
=
(\Re z-\Re a)\left[\frac{1}{|z-a|^2}+\frac{1}{|z-\bar a|^2}\right]\,.
$$
Note that $|F(z)|$ and $\frac{1}{|z-a|^2}+\frac{1}{|z-\bar a|^2}$ are symmetric with respect to the vertical line through $a$ and $\bar a$, while $\Re z-\Re a$ is antisymmetric, so in the full integral $\int_{\mathbb D} \frac{\Re F(z)}{|F(z)|}\,d\mu(z)$, the integrals over the yellow and the blue domain cancel out, so only the integral over the orange domain remains (see the figure).
However in the orange ddomain $\Re z-\Re a\le 0$, so, using the inequality $u^2+v^2\ge \frac 12(u+v)^2$ we can estimate the integral over it from above by the integral of $(\Re z -\Re a)\frac{G(z)^2}{|F(z)|}$ where $G(z)=\frac 1{|z-a|}+\frac 1{|z-\bar a|}$.
We shall combine this integral with the integral of $(\Re z-\Re a)\frac{|F(z)|}2$ (which can also be taken only over orange domain instead of the whole disk due to the same (anti)symmetry to get
$$
\int_{\text{orange}}(\Re z-\Re a)\left[\frac{|F(z)|}2+\frac{G(z)^2}{2|F(z)|}\right]\,d\mu(z)
\\
\le
\int_{\text{orange}}(\Re z-\Re a)G(z)\,d\mu(z)=
\int_{\mathbb D}(\Re z-\Re a)G(z)\,d\mu(z)
$$
(one more Cauchy-Schwarz and symmetry).
We are now left with the integral $\int_\mathbb D\frac{\Re z+\Re a}2|F(z)|\,d\mu(z)$. Our aim will be also to replace $|F(z)|$ by $G(z)$ in it, i.e., we want to show now that
$$
\int_\mathbb D(\Re z+\Re a)[G(z)-|F(z)|]\,d\mu(z)\ge 0\,.
$$
Notice that the integrand can be negative only to the left of the line through $-a$ and $-\bar a$. We shall show that on each horizontal line between $\Im a$ and $-\Im a$, the integral over the dangerous blue area (see the picture) is more than compensated by the integral over the yellow area (and the integral over the orange area is, of course, non-negative).
Since we clearly have $\Re z+\Re a\ge |\Re z^*+\Re a|$ for every $z$ in the yellow domain and the symmetric with respect to the imaginary axis to it point $z^*$ in the dangerous blue domain, it will suffice to show that
$$
G(z)-|F(z)|\ge G(z^*)-|F(z^*)|\,.
$$
Since
$$
|F(z)|=2\frac{|z-b|}{|z-a|\cdot|z-\bar a|}
$$
where $b=\Re a=\frac{a+\bar a}2$ and
$$
G(z)=\frac{|z-a|+|z-\bar a|}{|z-a|\cdot|z-\bar a|}\,,
$$
we see (taking into account obvious symmetries) that the inequality we need will follow from the following geometric statement:
Take a vertical segment of length $U$. Take a point $P$ on some horizontal line crossing this segment to the right of the segment but within the disk based on the segment as diameter. Then, if we move $p$ further from the segment along the same horizontal line (not necessarily staying in the disk), the expression
$$
\frac{A+B-2C}{AB}
$$
(see the figure) will become smaller.
By the standard elementary geometry formula for the median length (a.k.a. the "parallelogram identity"), we have $4C^2=2A^2+2B^2-U^2$, so
$$
\frac{A+B-2C}{AB}=\frac{(A+B)^2-4C^2}{AB(A+B+2C)}
\\
=\frac{(U^2-A^2-B^2)+2AB}{AB(A+B+2C)}
\\
=\frac{U^2-A^2-B^2}{AB(A+B+2C)}+\frac 2{A+B+2C}
$$
Now the second fraction obviously decreases all the way through while the first has a decreasing numerator that stays positive within the disk and negative outside and increasing positive denominator. This is more than enough to draw the desired conclusion. Most likely the whole expression continues to decrease beyond the disk, but I'm too lazy to investigate that. We have enough for our purposes as it is.
Now, bringing everything together and restoring the integration domain to the full disk adding the anti-symmetric regions as necessary, we see that the fancy integral we were interested in is bounded from above by the integral
$$
\int_{\mathbb D}\frac{3\Re z-\Re a}2G(z)\,d\mu(z)=
\int_{\mathbb D}\frac{3\Re z-\Re a}{|z-a|}\,d\mu(z)\,.
$$
It remains to compute this integral.
The one over the absolute value singularity at $a$ is just begging for switching to polar coordinates with the center at $a$. So, we will parametrize $z\in\mathbb D$ by $r$ and $\omega$ as on the figure below:
We have $\mathbb D=\{z(r,\omega):-\frac\pi 2\le\omega\le\frac\pi 2, 0\le r\le 2\cos\omega\}$, $|z-a|=r$, $3\Re z-\Re a=2\Re a-3r\sin(\omega+\gamma)=2\cos\theta-3r\sin(\omega+\gamma)$, $\gamma=\frac\pi 2-\omega$ and $d\mu(z)=r\,dr\,d\omega$. Thus our integral is
$$
\int_{-\pi/2}^{\pi/2}\left[\int_0^{2\cos\omega}(2\cos\theta - 3r\sin(\omega+\gamma))\,dr\right]\,d\omega
\\
=\int_{-\pi/2}^{\pi/2}
\left[(4\cos\theta\cos\omega - 6\sin(\omega+\gamma))\cos^2\omega\right]\,d\omega
\\
=
\int_{-\pi/2}^{\pi/2}
\left[(4\cos\theta\cos\omega - 6\sin\omega\cos\gamma\cos^2\omega-6\sin\gamma\cos^3\omega\right]\,d\omega
\\
=
4\cos\theta\int_{0}^{\pi/2}(2\cos\omega-3\cos^3\omega)\,d\omega
\\
=4\cos\theta\int_0^{\pi/2}(-\cos\omega+3\sin^2\omega\cos\omega)\,d\omega
\\
=4\cos\theta(-1+1)=0\,.
$$
We are done.
