6. Let $x, y \in \mathbf{R}$, and
$$
\log _{4}(x+2 y)+\log _{4}(x-2 y)=1 \text {. }
$$

Then the minimum value of $x-|y|$ is ( ).
(A) $\sqrt{3}$
(B) 2
(C) $2 \sqrt{3}$
(D) 4
6. A.

From the problem, we get $x^{2}-4 y^{2}=4$ and $x>2|y|$.
Assume
$$
x=2 \sec \theta, y=\tan \theta\left(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right) \text {. }
$$

Then $x-|y|=\frac{2}{\cos \theta}-\frac{|\sin \theta|}{\cos \theta}$.
Obviously, the above expression is an even function of $\theta$.
Let $t=\frac{2-\sin \theta}{\cos \theta}\left(\theta \in\left[0, \frac{\pi}{2}\right)\right)$.
Thus, $\sin \theta+t \cos \theta=2$
$\Rightarrow \sqrt{t^{2}+1} \sin (\theta+\varphi)=2$
$\Rightarrow \frac{2}{\sqrt{t^{2}+1}} \leqslant 1 \Rightarrow t^{2} \geqslant 3$.
When and only when $x=\frac{4}{\sqrt{3}},|y|=\frac{1}{\sqrt{3}}$, $t$ takes the minimum value
$\sqrt{3}$.