Let $r$ be sufficiently large (which you assume).
It is known that the $r$-th prime has size at least $r \log r$.
It is also known that the number of primes below $x$, $x$ large enough, is at least
$x/\log x$.

So the number of primes below your $p^2$ is at least

`\[\frac{(r \log r)^2}{ \log ( (r \log r)^2)} =\frac{(r \log r)^2}{ 2\log r + 2 \log \log r)}\ge \frac{r^2 \log r}{3} \ge 2 r^2 .\]`

Since by definition $p$ is the $r$-th prime one has
$2 r^2 - r \ge r^2$ in between.

To make the size condition explict and or sharpen the approach the results from Dusart's thesis will be useful.

Addition in view of the Edit. The argument above also yields a better bound of
$(1/2 - \epsilon) r^2 \log r$. Using similar results in the converse direction one can get $(1/2 + \epsilon) r^2 \log r$ as upper bound. I guess also an ansymptotic equality but I did not really check. So you have an extra logarithmic factor, which also explains why, as said by Noam Elkies, not even the full strength of PNT is needed to get what you claim.
As alluded to if you want to optimize this check the linked document for results.
But to exclude those belowe $p$ is essentially irrelevant as the contribution is minimal.