Here is an attempt for a relatively simple proof. I would be curious about your comments.
Case (a): $3r^2 + 8r- 4 > 0$ ($r > \approx 0.43$).
Then (I) is clearly increasing in $q$. But (II) implies that $q < \frac{r+2}{3}$. Hence, (I) can never attain a larger value than for $q = \frac{r+2}{3}$. Inserting this into (I), the condition
$\left(\frac{r+2}{3}\right)^2 (2-r) + \frac{r+2}{3} (3r^2 +8r-4) - 8r^2 > 0$ needs to be satisfied. But one can simplify this to
$\frac{8}{9} (1-r)^2 (r-2) > 0$, which can never hold for $0.
Case (b): $3r^2 + 8r - 4 < 0$ ($r <\approx 0.43$).
Then it is easy to see that (I) is a strictly convex function in $q$ over the relevant range, which attains a negative value for $q=0$, first decreases, and then increases. At best, there exists some critical threshold $\tilde{q} < 1$ starting from which the function becomes positive.
But in order for (II) to be satisfied, it cannot be the case that $q \geq \frac{r+2}{3}$. However, we know from case (a) that the function in (I) is certainly negative for $q = \frac{r+2}{3}$. Thus, also case (b) both inequalities cannot hold at the same time.