asymptotic for li(x)-Ri(x)

Yes, the stated asymptotics (and much more) is true. The idea is to truncate $\operatorname{Ri}(x)$ appropriately.

Let us use the series representation (see here) $$\operatorname{li}(t)=\gamma+\log\log t+\sum_{k=1}^\infty\frac{(\log t)^k}{k\cdot k!},\qquad t>1.$$ This implies $$\operatorname{li}(t)=\gamma+\log\log t+O(\log t),\qquad 1<t<e,$$ hence also $$\operatorname{li}(x^{1/n})=\gamma+\log\log x-\log n+O\left(\frac{\log x}{n}\right),\qquad n>\log x.$$ As a result, for $x>3$ we get \begin{align*}\sum_{n>\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})&=O(1)+\sum_{n>\log x}\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt] &=O\left((\log\log x)^2\right)+\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt] &=O\left((\log\log x)^2\right).\end{align*} Here we used that $$\sum_{n=1}^\infty\frac{\mu(n)}{n}=0\qquad\text{and}\qquad\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$ by the prime number theorem. To summarize so far, $$\operatorname{Ri}(x)=\sum_{n\leq\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})+O\left((\log\log x)^2\right),\qquad x>3.$$ The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get $$\operatorname{Ri}(x)=\operatorname{li}(x)-\frac{1}{2}\operatorname{li}(x^{1/2})+O\left(\frac{x^{1/3}}{\log x}\right),\qquad x>3.$$

Since we have $${\rm Ri}(x)=1+\sum_{k\ge 1}\frac{(\log x)^k}{k\cdot k!\zeta(k+1)},$$ which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by $$\sum_{n\ge 1}\frac{\mu(n)}{n}=0$$ we have
\begin{align} \lim_{x\rightarrow +\infty}\frac{{\rm li}(x)-{\rm Ri}(x)}{\frac{1}{2}{\rm li}(x^{1/2})}&=\lim_{x\rightarrow +\infty}\frac{\frac{1}{\log x}-\frac{1}{x\log x}\sum_{k\ge 1}\frac{(\log x)^k}{k!\zeta(k+1)}}{\frac{1}{2}\frac{1}{\log(x^{1/2})}\cdot\frac{1}{2}x^{-1/2}}\\ &=\lim_{x\rightarrow +\infty}\left(2\sqrt{x}\left(1-\frac{1}{x}\sum_{k\ge 1}\frac{(\log x)^k}{k!}\sum_{n\ge 1}\frac{\mu(n)}{n^{k+1}}\right)\right)\\ &=\lim_{x\rightarrow +\infty}\left(2\sqrt{x}\left(1-\frac{1}{x}\sum_{n\ge 1}\frac{\mu(n)}{n}(x^{1/n}-1)\right)\right)\\ &=1-2\lim_{x\rightarrow +\infty}\left(\frac{1}{\sqrt{x}}\sum_{n\ge 3}\frac{\mu(n)}{n}(x^{1/n}-1)\right). \end{align} On the other hand, for $x$ sufficiently large, $$\sum_{3\le n\le x}\frac{\mu(n)}{n}(x^{1/n}-1)\ll x^{1/3}\log x$$ and $$\sum_{n>x}\frac{\mu(n)}{n}(x^{1/n}-1)\ll \sum_{n>x}\frac{1}{n}\left(\exp\left(\frac{\log x}{n}\right)-1\right)\ll \sum_{n>x}\frac{\log x}{n^2}\ll 1.$$ Hence we get $$\lim_{x\rightarrow +\infty}\left(\frac{1}{\sqrt{x}}\sum_{n\ge 3}\frac{\mu(n)}{n}(x^{1/n}-1)\right)=0.$$ Which completes the proof.