Solution:

Let $NM = s$ and $SR = t$ be the side lengths of the two squares. By the Pythagorean theorem, we have $AB = 5\sqrt{13}$. From the above figure, triangles $NMB$, $ACB$ and $ARS$ are similar. Thus, $MB = \frac{BC \cdot NM}{AC} = \frac{2s}{3}$ and $AR = \frac{AC \cdot RS}{BC} = \frac{3t}{2}$. We see that $5\sqrt{13} = AB = AR + RQ + QM + MB = \frac{5s}{3} + \frac{5t}{2}$ or $\sqrt{13} = \frac{s}{3} + \frac{t}{2}$.

Hence, by Cauchy-Schwarz inequality, we get
$$\left(s^2 + t^2\right)\left(\frac{1}{3^2} + \frac{1}{2^2}\right) \geq \left(\frac{s}{3} + \frac{t}{2}\right)^2 = 13$$
so that $s^2 + t^2 \geq 13 / \left(\frac{1}{9} + \frac{1}{4}\right) = 36$. Hence, the minimum sum of the areas of the squares $S_1$ and $S_2$ is $36$, which is attained when $s = 12 / \sqrt{13}$ and $t = 18 / \sqrt{13}$.