Let $g:X\times S \rightarrow R $ be a twice continuously differentiable function. What conditions suffice to get $G(s)=\int_X g(x,s)dx$ is monotone in $s$?
Obviously an easy one would be if $g(x, s)$ is monotone in $s$ for all $x$ but I am hoping that there are much weaker conditions out there. Feel free to make additional assumptions such as $\frac{\partial^2 g(s,x)}{\partial s \partial x} \geq 0$ etc. though I am not sure how helpful that one is.
I'm assuming $S$ is an interval, possibly infinite, and that by "monotone" you mean nondecreasing. Of course, for nonincreasing you can reverse the inequalities.
Assuming $g(x,s)$ is absolutely continuous with respect to $s$ for every $x$, with $\partial g/\partial s \in L^1(X \times B)$ for each compact $B$ (to allow using Fubini's theorem), for $s_1 < s_2$ we have
$$ G(s_2) - G(s_1) = \int_X (g(x,s_2) - g(x,s_1))\; dx = \int_X \int_{s_1}^{s_2} \dfrac{\partial g}{\partial s}(x,s)\; ds\; dx =\int_{s_1}^{s_2} \int_X \dfrac{\partial g}{\partial s}(x,s)\; dx\; ds $$
so it suffices to have $\displaystyle \int_X \dfrac{\partial g}{\partial s}(x,s)\; dx \ge 0$ for all $s$. On the other hand, if $\displaystyle \int_{X} \dfrac{\partial g}{\partial s}(x,s) \; dx < 0$ on some interval $(s_1, s_2)$ we would have $G(s_2) < G(s_1)$.