I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$,
$$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big|
= \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$
The log of this quantity is bounded above, so 
it can not increase like $(1/2)\log(n)$. But it is not bounded below, it can takes large negative values for the integers $n$ such that $\alpha^n$ is close to $1$.  

ADDENDUM : one can check that 
$$\limsup_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} = 0,$$
whereas 
$$\liminf_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} \le -1.$$
Indeed, setting $\alpha = e^{i\theta}$ with $\theta \in \mathbb{R}$, Dirichlet principle ensures that $|\alpha^n-1| = |e^{in\theta}-1| \le \mathrm{dist}(n\theta,\mathbb{Z}) < 1/n$ for infinitely many $n$. 
Actually, the liminf above is $-1$ for almost every $\alpha$.