\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 $$f(t,x):=\sum_{k=1}^\infty\cos kt\, e^{-k x}=\frac{\cos t-e^{-x}}{2 (\cosh x-\cos t)}.$$ 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.
Also, $$f(t,x)=\frac{\sinh x}{2 (\cosh x-\cos t)}-\frac12.$$ So, by monotone convergence, $$\int_0^1 dx\,x^{r-1}f(t,x)\to\int_0^1 dx\,x^{r-1}\Big(\frac{\sinh x}{2 (\cosh x-1)}-\frac12\Big)=\infty$$ as $t\to2n\pi$ for any integer $n$, since $\sinh x\sim x$ and $\cosh x-1\sim x^2/2$ as $x\to0$. On the other hand, $$|f(t,x)|\le\sum_{k=1}^\infty e^{-k x}=e^{-x}$$ and hence for $r\in(0,1)$ we have $$\int_1^\infty dx\,x^{r-1}|f(t,x)|\le\int_1^\infty dx\,e^{-x}<1.$$ Thus, $$F_r(t)\to\infty$$ as $t\to2n\pi$ for any integer $n$.