\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. 

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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 $n$.