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It already suffices that $f$ and $g$ be even and nondecreasing on $[0,1]$ (which of course is the case if $f$ and $g$ are even and convex). Indeed, then the identity $|\sin|^2+|\cos|^2=1$ implies that for all real $u,v$ we have $$(|\sin u|-|\sin v|)(|\cos u|-|\cos v|)\le0$$ and hence $$\begin{aligned}h(u,v)&:=[f(\sin u)-f(\sin v)][g(\cos u)-g(\cos v)] \\ &=[f(|\sin u|)-f(|\sin v|)][g(|\cos u|)-g(|\cos v|)]\le0, \end{aligned}$$ so that the difference between the left-hand side of your inequality and its right-hand side is $$\int_1^\infty\int_1^\infty\frac{du\,dv\,h(u,v)}{u^2 v^2}\le0. $$