Take any $t\in(0,2\pi)$. Then \begin{equation*} \begin{aligned} 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{aligned} \tag{1} \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 convergence of the series defining $F_r(t)$, as well as interchange of the summation and the integration, are justified at the end of this answer.)
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, for $r\in(0,1)$ we have $$\int_0^1 dx\,x^{r-1}f(t,x)\to\int_0^1 dx\,x^{r-1}\Big(\frac{\sinh x}{2 (\cosh x-1)}-\frac12\Big)=\infty$$ 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, for real $x>1$, $$|f(t,x)|\le\sum_{k=1}^\infty e^{-k x}\le 2e^{-x}$$ and hence for $r\in(0,1)$ we have $$\int_1^\infty dx\,x^{r-1}|f(t,x)|\le2\int_1^\infty dx\,e^{-x}<1.$$ Thus, $$\Gamma(r)F_r(t)\ge\int_0^1 dx\,x^{r-1}f(t,x)-\int_1^\infty dx\,x^{r-1}|f(t,x)|\to\infty$$ as $t\to2n\pi$, for any integer $n$. So, the claims about the limits follow as well.
(Justification of the convergence of the series defining $F_r(t)$, as well as of the interchange of the summation and the integration in (1).) Take any real $r\in(0,2)$. Take any $t\in(0,2\pi)$. Let us show that the limit \begin{equation*} F_r(t)=\lim_{n\to\infty}F_{r,n}(t) \end{equation*} exists, where \begin{equation*} F_{r,n}(t):=\sum_{k=1}^n\frac{\cos kt}{k^r}. \end{equation*}
Indeed, for all natural $m$ and $n>m$, \begin{equation*} \begin{aligned} F_{r,n}(t)-F_{r,m}(t)&=\sum_{k=m+1}^n\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=m+1}^n\cos kt\, e^{-k x}. \end{aligned} \end{equation*} Next, for $x>0$, \begin{align*} \Big|\sum_{k=m+1}^n\cos kt\, e^{-k x}\Big| & =\Big|\Re\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ & \le\Big|\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ &=e^{-(m+1)x}\,\frac{|1-e^{-(n-m)(it+x)}|}{|1-e^{-it-x}|} \\ &\le e^{-(m+1)x}\,\frac2{2\sin(t/2)\,e^{-x/2}} \\ &\le \,\frac{e^{-mx}}{\sin(t/2)}. \end{align*} So, $F_{r,n}(t)$ is Cauchy-convergent in $n$ and hence convergent in $n$. Also, by dominated convergence, we can interchange the summation and the integration in (1).