$\newcommand{\la}{\lambda}\newcommand{\Ga}{\Gamma}$Write 
\begin{equation}
	I(\la)=\int_0^\infty dx\,f(x)J(\la x),
\end{equation}
where 
\begin{equation}
	J(z):=\int_0^1 dt\,e^{i z t^a}=\frac b{z^b}K(z),
\end{equation}
\begin{equation}
	K(z):=\int_0^z du\,u^{b-1}e^{i u}, 
\end{equation}
$b:=1/a\in(0,1)$. 
Note that 
\begin{equation}
	K(z)\to\int_0^\infty du\,u^{b-1}e^{i u}=(-i)^b\Ga(b)  
\end{equation}
as $z\to\infty$. 
Also, since $u^{b-1}$ decreases to $0$ as $u$ increases from $0$ to $\infty$, we have $|K(z)|\le C$ for some real $C>0$ and all real $z\ge0$. 

So, letting $\la\to\infty$, by dominated convergence we get  
\begin{equation}
\la^b	I(\la)=\int_0^\infty dx\,f(x)\frac b{x^b}K(\la x) \\ 
\to (-i)^b\Ga(b+1)\int_0^\infty dx\,f(x)/x^b,
\end{equation} 
so that 
\begin{equation}
	I(\la)\sim(i\la)^{-b}\,\Ga(b+1)\int_0^\infty dx\,f(x)/x^b. 
\end{equation}