Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write
$$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$
then we have, using Riemann-Stieltjes integrals,
$$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ 
$$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 
+ O( \ln x \sup_{x < t \leq 2x} R(t) ). $$
If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation
$$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$
we have
$$ \hat F(s) = \int_{2-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$
In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. By a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory) we infer the two-sided estimates
$$ F(x)-\ln 2 = \Omega_\pm(x^{\sigma-1}). $$
This implies, by our initial calculation,
$$ R(x) = \Omega_\pm(x^{\sigma-1}/\ln x). $$
Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$, while in the case of a multiple root for $\zeta(s)$ we can improve the bounds by the appropriate power of $\ln x$.