Solution:
First, note that the polynomial $x^{4} - 10x^{2} + 1$ has both $\sqrt{3} + \sqrt{2}$ and $\sqrt{3} - \sqrt{2}$ as roots. It suffices to check whether a polynomial of degree at most 3 belongs in $\mathcal{S}$. Suppose $f(x) = a x^{3} + b x^{2} + c x + d \in \mathcal{S}$. We compute

$$(\sqrt{3} + \sqrt{2})^{3} - (\sqrt{3} - \sqrt{2})^{3} = 22\sqrt{2}$$
$$(\sqrt{3} + \sqrt{2})^{2} - (\sqrt{3} - \sqrt{2})^{2} = 4\sqrt{6}$$
$$(\sqrt{3} + \sqrt{2})^{1} - (\sqrt{3} - \sqrt{2})^{1} = 2\sqrt{2},$$

so we get that

$$f(\sqrt{3} + \sqrt{2}) - f(\sqrt{3} - \sqrt{2}) = (22\sqrt{2})a + (4\sqrt{6})b + (2\sqrt{2})c.$$

By resolving linear dependencies, it's clear that $b = 0$ and $c = -11a$. It follows that if $f$ is not the zero polynomial, it must be cubic. It is then clear that $f(x) = x^{3} - 11x + d$ has minimal degree in $\mathcal{S}$, and thus $Q(10) - Q(0) = f(10) - f(0) = \boxed{890}$.