$\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}$If a function $h\colon\R\to\R$ is monotonic and does not change sign on an interval $[a,b]$, then it is easy to see (make a picture) that for all real $s$ 
\begin{equation*}
	\Big|\int_{[a,b]}h(u)\sin(u+s)\,du\Big|\le3\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du,
\end{equation*}
whence 
\begin{equation*}
	\Big|\int_{[a,b]}h(u)e^{iu}\,du\Big|\le6\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du. 
\end{equation*}
Since $x^\al/F'(x)$ is monotonic in $x\in[1,T]$, $F'$ cannot change its (nonzero) sign on $[1,T]$. Making now the substitution $u=F(x)$, letting here  
\begin{equation*}
	h(u):=\frac{g(u)^\al}{F'(g(u))} \text{ with } g:=F^{-1},  
\end{equation*} 
and then making the inverse substitution $x=g(u)$, we get 
\begin{equation*}
\begin{aligned}
	I_\al(T)=
	&\Big|\int_{F([1,T])}h(u)e^{iu}\,du\Big| \\
	&\le6\sup_{c\in\R}\int_{F([1,T])\cap[c,c+\pi]}|h(u)|\,du \\ 
	&=6\sup_{c\in\R}\int_{[1,T]\cap g([c,c+\pi])}x^\al\,dx. 
\end{aligned}
\tag{1}
\end{equation*}

Recall that $F'$ cannot change its (nonzero) sign on $[1,T]$. 

If $F'>0$ on $[1,T]$, then for any $x,y$ in $[1,T]$ such that $x\le y$ 
\begin{equation*}
	|F(y)-F(x)|=F(y)-F(x)\ge\frac{(y-x)^2}{2T}, 
\end{equation*}
since $F''\ge1/T$ and $F'>0$. So, 
\begin{equation*}
	|g(c)-g(c+\pi)|\le\sqrt{2\pi T}, 
\end{equation*}
and hence, by (1), 
\begin{equation*}
	I_\al(T)
	\le6\int_0^{\sqrt{2\pi T}}x^\al\,dx
=	\frac C{1+\al}\, T^{(1+\al)/2},
\end{equation*}
whence $C$ is a universal positive real constant. 

The latter bound is indeed an improvement of the corresponding bound in your post. 

Morever, the latter bound is optimal, as it is asymptotically attained (up to a positive real constant depending only on $\al$) if $F(x)=\dfrac{x^2}{2T}$: 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/5ySt4.png