Well, there are two contributions because the exponent is zero at $t=0$ and $t=\pi/2$. Let's consider $t=0$ first. In the immediate neighborhood of $t=0$ (we'll get to what that means in a bit), the exponent behaves as $x t^3$ so that as $x\to\infty$, we have
$$I(x) \sim \int_0^{\infty} dt \, e^{-x t^3} = \frac{\Gamma\left ( \frac{4}{3}\right )}{x^{1/3}} = \frac{\Gamma\left ( \frac{1}{3}\right )}{3 x^{1/3}} \quad (x\to\infty)$$
What is the size of the neighborhood? Well, we want $0 \lt x t^3 \lt \epsilon$ for some small $\epsilon$, so we have $0 \lt t \lt (\epsilon/x)^{1/3}$.
We also have a contribution in a neighborhood near $t=\pi/2$; we Taylor expand and get that
$$t^3 \cos{t} = -\frac{\pi^3}{8} \left ( t-\frac{\pi}{2}\right ) + O\left [ \left ( t-\frac{\pi}{2}\right )^2\right ]$$
Really, we are only interested in positive values of this expansion, as the exponent is positive through the integration region. If we look at only an immediate neighborhood near $t=\pi/2$, but with $t \lt \pi/2$, then we may approximate the contribution to the integral there as
$$\int_0^{\infty} dy \, e^{-\pi^3 x y/8} = \frac{8}{\pi^3 x}$$
It doesn't make sense to simply add these two terms together and declare them the leading behavior of $I(x)$ until we investigate the next leading behavior of the contribution at $x=0$. Note that the next contribution in the exponent is $x t^5/2$; within the interval of interest, this is $O(x^{-2/3})$, so we may Taylor expand this exponential term separately. The result is the following integral for the next contribution at $t=0$:
$$\frac12 x\int_0^{\infty} dt \;t^5 \, e^{-x t^3} = \frac1{6 x}$$
Note that this is $O(1/x)$ as is the leading contribution from $t=\pi/2$, so we may add these. The stated result follows.
Best Answer
The substitution $t=x(1-y)^{1/3}$, and Watson's lemma, give $$I(x)=\frac{x}{3}e^{x^3}\int_0^1(1-y)^{-2/3}e^{-x^3 y}\,dy\asymp\frac{e^{x^3}}{3}\sum_{n=0}^{(\infty)}\frac{\Gamma(n+2/3)}{\Gamma(2/3)x^{3n+2}}=e^{x^3}\left(\frac{1}{3x^2}+\frac{2}{9x^5}+\ldots\right).$$