Borwein, Bradley, and Crandall state on page 249 here that
... it should be the case that, in some appropriate sense
$$ \pi(x) \sim \text{Ri}(x) - \sum_{\rho} \text{Ri}(x^\rho) \tag{4} $$
with $\text{Ri}$ denoting the Riemann function defined:
$$ \text{Ri}(x) = \sum_{m=1}^\infty \frac{\mu(m)}{m}\text{li}(x^{1/m}). \tag{5} $$
This relation (4) has been called “exact” [94], yet we could not locate a proof in the literature; such
a proof should be nontrivial, as the conditionally convergent series involved are problematic. In any case relation (4) is quite accurate ...
So it appears we don't really know if this relation is indeed true or not.
But as the quote says, it does appear this relation at least provides a good estimate of $\pi$. In fact $\text{Ri}(x)$ alone provides a good estimate of $\pi(x)$. For example, $\pi(10^{20}) = 2220819602560918840 $, and here's $\text{Ri}(10^{20})$ evaluated in Mathematica:
In[186]:= Floor[RiemannR[10^20]]
Out[186]= 2220819602556027015
This gives a relative error of about $2.2 \cdot 10^{-12}$, meaning the first 11 digits are correct!
Now how about incorporating the zeros $\rho$? Well they actually seem to make things worse (at least for a 'small' number of zeros). I took the first $14400$ zeros of $\zeta$, to a precision of 30 digits, and got an answer with relative error $3.1 \cdot 10^{-7}$. In fact the more zeros I chose, the worse the relative error became.

So to answer your question, I think this formula seems to provide an excellent approximation for $\pi(x)$. However, at the end of the day we'll only be able to get an approximation, not an exact answer.