Here is a full characterization.
Theorem. The function $e^{-|x|^a}$ is positive definite for $0\le a \le 2$ and is not positive definite for $a>2$. Thus, its Fourier transform is positive and non positive for the said ranges.
Proof. The claim $a=0$ is trivial. Assume $0<a<2$. Then, it can be shown that \begin{equation*} -|x|^a = C_a\int_{-\infty}^\infty \frac{\cos (xt) - 1}{|t|^{a+1}}dt, \end{equation*} where $C_a$ is a constant depending on $a$. Since $\cos(xt)$ is a positive definite function, we see that $-|x|^a$ is conditionally positive definite (because of the $-1$ term). Hence, $\exp(-|x|^a)$ is positive definite, and consequently by Bochner's theorem, it's FT is positive.
For $a>2$, it is easy to construct numerical examples where the associated function is not positive definite, and hence its FT is not positive. Carlo Beenakker's answer gives an example.
To obtain a formal proof of this, here's an outline by contradiction. In particular, suppose that for some $a > 2$, the kernel $e^{-|x-y|^a}$ is positive definite, $|x-y|^a$ is negative definite. Thus, by appealing to Schoenberg's theorem, it must be the case that $d(x,y) := |x-y|^{a/2}$ is a metric on $\mathbb{R}$. Choosing $x,y,z=(0,1,2)$ and comparing $d(x,y)=d(y,z)=1$ but $d(x,z)=2^{a/2}>2$, a contradiction to the triangle inequality.
Reference. Chapter 5, Positive definite matrices, R. Bhatia. Princeton University Press, 2007.