We have the following Lemma, which is related to the properties of the Poisson kernel:
$$ \text{if }a\in(0,1),\qquad \int_{0}^{2\pi}\log\left|a-e^{it}\right|\,dt = 0,$$
$$ \text{if }a>1,\qquad \int_{0}^{2\pi}\log\left|a-e^{it}\right|\,dt = 2\pi\log a.$$
Let us focus on $\iint_{B(0,R)}\log|x-y|\,d\mu$.
By symmetry we may clearly assume that $x\in(0,R)$. With such assumption we have
$$ \iint_{B(0,R)}\log|x-y|\,d\mu = \int_{0}^{2\pi}\int_{0}^{R}\rho \log|x-\rho e^{i\theta}|\,d\rho\,d\theta\\=\int_{0}^{R}2\pi\rho\log\rho+2\pi\rho\max\left(0,\log\frac{x}{\rho}\right)d\rho $$
hence the LHS equals
$$ \frac{\pi R^2}{2}(2\log R-1)+2\pi\int_{0}^{x}\rho\log\frac{x}{\rho}\,d\rho=\frac{\pi R^2}{2}(2\log R-1)+\frac{\pi x^2}{2}. $$
Similarly,
$$ \iint_{B(0,R)}\log\left|x-\frac{R^2 y}{\|y\|^2}\right|\,d\mu=\int_{0}^{R}\int_{0}^{2\pi}\rho \log\left|x-\frac{R^2}{\rho} e^{i\theta}\right|\,d\theta\,d\rho =\frac{\pi R^2}{2}(2\log R+1)$$
does not really depend on $x$. It follows that for any $R>0$ the minimum of the given integral, $f(x)=-\pi R^2+\frac{\pi}{2}\|x\|^2$, is attained at the origin, and it equals
$$ \iint_{B(0,R)}\log\frac{\|y\|^2}{R^2}\,d\mu = -\pi R^2. $$
