$|1-e^{i\theta}|=2\sin{\frac{\theta}{2}}$, so we need to evaluate $I=\int_0^{2\pi}\log |1-e^{i\theta}|d\theta$ as then the answer is $2\pi\log R+I$.
Now $\log|1-z|=\Re{\log(1-z)}$ is harmonic inside the unit disc, so by the mean value theorem for harmonic function, $\int_0^{2\pi}\log |1-re^{i\theta}|d\theta=0$ for any $r<1$.
But $\log|1-e^{i\theta}|$ is obviously integrable on the unit circle (since near $0$, $\log 2\sin{\frac{\theta}{2}}-\log \theta$ is continuos and the latter is clearly integrable, while near $2\pi$ we can use $\sin{\frac{\theta}{2}}=\sin{\frac{2\pi-\theta}{2}}$ and the result at zero) and $\log|1-re^{i\theta}| \to \log|1-e^{i\theta}|$ a.e.
But now if say $0 \le \theta \le \frac{\pi}{100}$ or $0 \le 2\pi-\theta \le \frac{\pi}{100}$, by drawing the perpendicular from $1$ to the $\theta$ ray which has length $|\sin \theta|$ it follows from elementary geometry that $|1-re^{i\theta}| \ge |\sin \theta|$, while for the rest $|1-re^{i\theta}| \ge c >0$ and since $|1-re^{i\theta}| \le 2$, we get that $|\log |1-re^{i\theta}|| \le \max {(\log 2, \log^- c-\log |\sin \theta|)}$ and that is integrable on the unit circle as before, hence we can apply the Lebesgue dominated convergence and conclude that $0=\int_0^{2\pi}\log |1-re^{i\theta}|d\theta \to I$, so $I=0$ and the final answer is $2\pi\log R$
As an aside, there is also a classic real variables proof that $I=0$ using the doubling formula for the $\sin$ and various changes of variable while the above can be expressed in terms of contour integrals if one wishes using $f(z)=\frac {\log (1-z)}{z}$ which is analytic inside the unit disc, or treat the original integral using $f(z)=\frac {\log R(1-z)}{z}$ which has $\log R$ as residue at $0$ etc, but of course it may not quite be what OP had in mind.
(Edit) As asked let's quickly use Cauchy rather than Poisson:
$I_1=\int_0^{2\pi}\log (2R\sin{\frac{\theta}{2}})d\theta=\int_0^{2\pi}\log R|1-e^{i\theta}|d\theta= \int_0^{2\pi} \Re {\log R(1-e^{i\theta})}d\theta=\Re {\int_0^{2\pi} \log R(1-e^{i\theta})d\theta}$
with the last equality holding because $d\theta$ is a real (positive) measure.
But now with the usual $e^{i\theta}=z, d\theta=\frac{1}{iz}dz$ we have;
$I_1=\Re {\int_{|z|=1} \frac{\log R(1-z)}{iz}dz}$
Now by Cauchy ${\int_{|z|=1} \frac{\log R(1-rz)}{iz}dz}= 2\pi \log R$ as the residue at zero of the integrand is $\frac{\log R}{i}$, while it is analytic anywhere else on the closed unit disc when $0 < r <1$. So we need to be able to pass to the limit $ r \to 1$ to conclude that the above unit circle integral (and hence its real part) is $2\pi \log R$ and the same argument as above works since the only problem comes from $\log |1-rz|$ near the boundary as everything else is obviously bounded so the same estimates work to show that we can use the Lebesgue dominated convergence and conclude that $I_1= 2\pi \log R$
(for $|z|=1$ we have $|\frac{\log R(1-rz)}{iz}|\le |\log R|+ |\log (1-rz)| \le |\log R|+ |\Re \log (1-rz)|+ |\Im \log (1-rz)| $ and $|\Re \log (1-rz)|=|\log |1-rz||$ as above, while $|\Im \log (1-rz)|=|\arg (1-rz)| \le \frac{\pi}{2}$ since $\Re (1-rz) >0, |z| =1$)
Best Answer
Let us define
$$I(a) := \int_0^\infty \frac{\ln(x)}{x^2+a^2}dx = \frac{\pi \ln(a)}{2a}$$
As suggested in the comments by Yuriy S., we can use Leibniz' rule of differentiation under the integral sign here.$^{(1)}$ Take the derivative with respect to $a$ throughout:
$$\begin{align} I'(a) &= \int_0^\infty \ln(x) \cdot \frac{\partial}{\partial a} \left( \frac{1}{x^2+a^2} \right)dx = -2a \int_0^\infty \frac{\ln(x)}{(x^2+a^2)^2}dx \\ &= \frac \pi 2 \cdot \frac{d}{da} \left( \frac{\ln(a)}{a} \right) = \frac{\pi}{2a^2} \left(1 - \ln(a) \right) \end{align}$$
The integral you desire is given by $I'(e)$, divided by a factor of $-2e$, but that factor doesn't matter in the end since $I'(e) = 0$ and thus so does your integral.
$^{(1)}$ - Leibniz's rule more generally states the following:
$$ \frac{d}{dx} \int_a^b f(x,t)dt = \int_a^b \frac{\partial}{\partial x} f(x,t)dt$$
when $f$ and $f_x$ are continuous functions (over, if necessary, just some region of the $xt$ plane, e.g. for $t \in (a,b)$ and $x$ in some interval about the point where you intend to evaluate it). I believe it's sufficiently clear that these hold in your case.
A PDF detailing the rule and many examples can be found here.
I don't claim this to be the only way to evaluate the integral. Indeed, that it is equal to zero makes me wonder if there might be some means of transforming it so that you have an odd integrand on a symmetric interval. I'm also not 100% sure if this is the kind of method you wanted to look at given your mentions of complex contour integration.