Answer: 2014.
By condition, $rp^3 = -(p^2 + p) + 2rq^2 + (q^2 + q)$, and $p^2 + p$, $q^2 + q$, $2rq^2$ are even for all natural $p$ and $q$, then $rp^3$ is also even. Therefore, since $r$ and $p$ are prime, we see that either $p = 2$ or $r = 2$.

If $p = 2$, then the initial equality has the form $8r + 4 + 2 = 2rq^2 + q^2 + q$. But this is impossible because the right-hand side of this expression is greater than its left-hand side. Indeed, since $p = 2$ and $q \neq p$ is prime, we have $q > 2$, so $2rq^2 + q^2 + q > 8r + 4 + 2$.

Thus $r = 2$ and the initial equality has the form $2p^3 + p^2 + p = 4q^2 + q^2 + q$, or
$$p(2p^2 + p + 1) = q(5q + 1). \quad (1)$$
Since $p$ and $q$ are distinct prime numbers, we see that $p$ and $q$ are coprime. Then from (1) it follows that $2p^2 + p + 1$ is divisible by $q$, and $5q + 1$ is divisible by $p$. Therefore $2p^2 + p + 1 = mq$ and $5q + 1 = mp$ for some $m \in \mathbb{N}$. We put $q = (mp - 1)/5$ into the former equality, then
$$10p^2 + (5 - m^2)p + (m + 5) = 0. \quad (2)$$
Consider this equality as a quadratic equation with respect to $p$. Since the coefficients of (2) are integer and its root $p$ is integer, we obtain that the discriminant of (2) is necessarily a perfect square, i.e.
$$D = (m^2 - 5)^2 - 4 \cdot 10 \cdot (m + 5) = m^4 - 10m^2 - 40m - 175 = n^2 \quad (3)$$
for some nonnegative integer $n$. Since $D < m^4 - 175$, we see that for $m = 1, 2, 3$ the discriminant is negative. Moreover, for $m = 4$ we have $D = 256 - 160 - 160 - 175 < 0$. Therefore $m \ge 5$.

From (3) it follows that $(m^2 - 5)^2 > n^2$. Show that $n^2 > (m^2 - 11)^2$. If not, then from (3) it follows that
$$(m^2 - 5)^2 - 40(m + 5) \le (m^2 - 11)^2 \Leftrightarrow 3m^2 - 10m - 74 \le 0. \quad (4)$$
It is easy to see that the latter inequality holds only for $m \le 6$. But $m \ge 5$, so to prove that $n^2 > (m^2 - 11)^2$ it remains to show that for $m = 5$ and $m = 6$ equality (2) does not hold for prime $p$.

If $m = 5$, then (2) has the form $10p^2 - 20p + 10 = 0$ and has exactly one root $p = 1$; but $1$ is not prime number. If $m = 6$, then the discriminant of (2) is equal to $D = 1296 - 360 - 240 - 175 = 521$; but $521$ is not perfect square.

Therefore $(m^2 - 11)^2 < n^2 < (m^2 - 5)^2$, i.e., taking into account that $m \ge 7$ and $n \ge 0$, we have $m^2 - 11 < n < m^2 - 5$. Note that from (3) it follows that $m$ and $n$ have different parity, so $n$ can admit only two values $n = m^2 - 9$ or $n = m^2 - 7$.

If $n = m^2 - 9$, then from (3) we obtain $m^4 - 10m^2 - 40m - 175 = (m^2 - 9)^2$ or $m^2 - 5m - 32 = 0$. However, it is easy to see that the discriminant of this equation is equal to $153$, so the equation has no integer solutions.

If $n = m^2 - 7$, then from (3) we get $m^4 - 10m^2 - 40m - 175 = (m^2 - 7)^2$ or $m^2 - 10m - 56 = 0$. We see that $m = -4$ and $m = 14$ are the roots of this equation. Since $m \in \mathbb{N}$, we have $m = 14$.

Therefore $m = 14$, then $n = m^2 - 7 = 189$. From (2) we obtain $p = (m^2 - 5 \pm n)/20 = (191 \pm 189)/20$. Since $p$ is integer, we have $p = 19$ which is prime. Hence $q = (mp - 1)/5 = (14 \cdot 19 - 1)/5 = 265/5 = 53$.

Thus the required value of the product is $pqr = 19 \cdot 53 \cdot 2 = 2014$.