If you are not interested in a particularly elegant expression, then you can very easily and simply calculate the CDF of $D$ as follows:
$$
{\rm P}(D \le s) = \frac{1}{{4\pi }}\int_{[ - 1,1] \times [0,2\pi ]} {\mathbf{1}\big(\sqrt {1 - x^2 \sin ^2 \theta } - x\cos \theta \le s\big)\,dx d\theta } ,\;\; 0 \leq s \leq 2,
$$
where $\mathbf{1}$ denotes the indicator function. Note that $D$ is supported on the set $[0,2]$ (hence the restriction $0 \leq s \leq 2$ above). Indeed, on the one hand, $D \geq 0$ since
$$
\sqrt {1 - x^2 \sin ^2 \theta } \ge x\cos \theta
$$
for any $x \in [-1,1]$ and $\theta \in [0,2\pi]$ (this is trivial if $x\cos\theta \leq 0$; otherwise take squares on both sides), and, on the other hand, $D \leq 2$ since
$$
\sqrt {1 - x^2 \sin ^2 \theta } - x\cos \theta \le 1 + 1 = 2
$$
(for $x$ and $\theta$ as above). Further note that the choices $(x,\theta)=(1,0)$ and $(x,\theta)=(-1,0)$ correspond to $D=0$ and $D=2$, respectively.
Finally, it should be noted that the double integral above can be calculated very quickly and accurately. For example,
it took about a second to obtain the approximation
$$
{\rm P}(D \leq 1) \approx 0.5813759999978363,
$$
indicating that ${\rm P}(D \leq 1) = 0.581376$, whereas Monte Carlo simulations ($10^7$ repetitions) yielded a much less accurate approximation
$$
{\rm P}(D \leq 1) \approx 0.5813805,
$$
in about $40$ seconds.