I think I found a way to apply the method of stationary phase (please give feedback if you think that the answer is not sound).
As Jon noted via partial integration the integral can be brought into the form $I(\alpha) = \alpha \int_0^\infty\!dx\,x^2 K_1(x) \sin[2 \alpha K_0(x)].$ Next I perform the substitution $y=2K_0(x)$ (note that $K_0$ is a monotonously falling function such that the inverse $K_0^{-1}$ is uniquely defined). I get $ I(\alpha) = \frac{\alpha}2 \int_0^\infty\!dy\,\sin(\alpha y)[K_0^{-1}(y/2)]^2; $ note that $K_1$ cancelled with the derivative of $K_0^{-1}$ via inverse function theorem and the fact that $K_0'(x)=-K_1(x)$.
This integral is almost ready for a stationary phase analysis, we replace $\sin(\alpha y) = \operatorname{Im} e^{i\alpha y}$ and note that the path of stationary phase starting from $y=0$ is along the imaginary axis. Thus, we substitute $y=i \zeta$ and get $I(\alpha)=\frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \operatorname{Re}[K_0^{-1}(i\zeta/2)]^2 ;$ here, we have analytically continued $K_0^{-1}$ into the complex plane.
The integral is dominated at $\zeta\approx 0$. Thus we can Taylor-expand $K_0^{-1}(i\zeta)$. We have $K_0(x) \sim e^{-x}$ thus $K_0^{-1}(y) \sim -\log y$. Using this asymptotic relation, we obtain $I(\alpha)\sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \overbrace{\operatorname{Re}[\log(i\zeta/2)]^2}^{\sim \log^2 (\zeta/2)} \sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta}\log^2 (\zeta/2) \sim \tfrac12 \log^2\alpha. $
I guess for the next term one has to work a bit harder and use $K_0^{-1}(y) \sim - \log y - \tfrac12 \log(-\log y)$. However, in principle it should be possible to extend this approach to next to leading terms.