\begin{equation}
F_r(t)=\sum_{k=1}^\infty\cos kt\, \frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}e^{-k x} \\ 
=\frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}\sum_{k=1}^\infty\cos kt\, e^{-k x} \\ 
=\frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}f(t,x),
\end{equation}
where 
$$f(t,x):=\frac{-\cos t-\sinh x+\cosh x}{2 (\cos t-\cosh x)}.$$
The partial derivative of $f(t,x)$ in $t$ is 
$$-\frac{\sin t \sinh x}{2 (\cos t-\cosh x)^2},$$
whose sign for $x>0$ is opposite to the sign of $\sin t$. So, the desired monotonicity follows.