Does $\sum_\rho R(x^\rho) \sim \sqrt{x}/\log(x)$ assume the Riemann hypothesis?

Question

I am learning about the exact formula for the prime counting function $\pi(x)=R(x)-\sum_{\rho}R(x^\rho)$ where $R$ is Riemann's R-function $R(x)=\sum_{k=1}^\infty\frac{\mu(k)}{k}li(x^{1/k})$, $li$ the logarithmic integral, and $\{\rho\}$ is the collection of all the zeros of the zeta function (trivial, non-trivial). I have read through Riesel and Gohl's article where they show that for the zeros on the critical line, the contributions of conjugate pairs of zeros yields an oscillatory function whose amplitude grows $\sqrt{x}/\log(x)$. Moreover, Wikipedia states the amplitude of $\sum_\rho R(x^\rho)$ grows as $\sqrt{x}/\log(x)$, presumably from the same source. My question is: does the Wikipedia statement assume the Riemann hypothesis? It seems that if there were a non-trivial zero off the critical line that this "noisy" term would grow faster? Is there any way to prove $\sum_\rho R(x^\rho)$ grows as $\sqrt{x}/\log(x)$, where the sum is taken over all the zeros?