Let $\mathcal{R}$ denote the set of Ramanujan primes, $\pi_{\mathcal{R}}(x)$ the number of Ramanujan primes less than $x$, and $$\theta_{\mathcal{R}}(x)=\sum_{\begin{array}{c}
p\leq x\\
p\in \mathcal{R}
\end{array}}\log p.$$ Then, writing this as a Riemann-Stieltjes integral, this will be $$\sum_{\begin{array}{c}
p\leq x\\
p\in\mathcal{R}
\end{array}}\left\{ \frac{x}{p}\right\} =\int_{2}^{x}\frac{1}{\log t}\left\{ \frac{x}{t}\right\} d\theta_{\mathcal{R}}(t).$$ To understand this quantity, we need to understand the number of Ramanujan primes less than $x$. The $n^{th}$ Ramanujan prime is the smallest integer $R_{n}$ such that $\pi(x)-\pi(x/2)\geq n$ for all $x\geq R_{n}$. Now, since $\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{\log^{2}x}\right),$ it follows that $$\pi(x)-\pi(x/2)=\frac{x}{2\log x}+O\left(\frac{x}{\log^{2}x}\right).$$ The number of Ramanujan primes less than $x$, $\pi_{\mathcal{R}}(x)$, is equal to the largest $n$ such that $\pi(y)-\pi(y/2)\geq n$ for all $y\geq x$. By the above asymptotics, it follows that $$n=\pi_{\mathcal{R}}(x)=\frac{x}{2\log x}+O\left(\frac{x}{\log^{2}x}\right),$$ and by partial summation, we have that $$\theta_{\mathcal{R}}(x)=\frac{x}{2}+O\left(\frac{x}{\log x}\right).$$ Hence, by using integration by parts and bounding error terms in the partial summation above, we have that $$\sum_{\begin{array}{c}
p\leq x\\
p\in\mathcal{R}
\end{array}}\left\{ \frac{x}{p}\right\} =\frac{1}{2}\int_{2}^{x}\frac{1}{\log t}\left\{ \frac{x}{t}\right\} dt+O\left(\frac{x}{\log^{2}x}\right).$$ In other words, $$\sum_{\begin{array}{c}
p\leq x\\
p\in\mathcal{R}
\end{array}}\left\{ \frac{x}{p}\right\} =\frac{1}{2}\sum_{p\leq x}\left\{ \frac{x}{p}\right\} +O\left(\frac{x}{\log^{2}x}\right).$$