Let $z=x+iy,$ where $x,y\in\mathbb R$.

Given that the real part of $\frac{z-2i}{z+2i}$ is zero.

Substitute $z=x+iy$: $\frac{z-2i}{z+2i}=\frac{x+i(y-2)}{x+i(y+2)}.$

Rationalize the denominator:

Expand the numerator: \begin{align*} (x+i(y-2))(x-i(y+2)) &=x^2-ix(y+2)+ix(y-2)+ (i)(-i)(y-2)(y+2) \ &=x^2 + x[-i(y+2)+i(y-2)] + (y^2-4) \ &=x^2 -4ix + y^2-4. \end{align*}

Hence, $\frac{z-2i}{z+2i}=\frac{x^2+y^2-4-4ix}{x^2+(y+2)^2}.$

Its real part is $\frac{x^2+y^2-4}{x^2+(y+2)^2}.$ This is given to be zero, so $x^2+y^2-4=0.$

Therefore, $x^2+y^2=4.$

So the locus of $z$ is the circle $|z|=2,$ centered at the origin with radius $2$.

We need the maximum value of $|z-(6+8i)|.$

This is the distance between a point $z$ on the circle $|z|=2$ and the fixed point $(6,8)$ in the Argand plane.

The distance from the origin to $6+8i$ is $|6+8i|=\sqrt{6^2+8^2}=\sqrt{36+64}=10.$

For a circle centered at the origin with radius $2$, the maximum distance from the fixed point at distance $10$ from the center is $10+2=12.$

Hence, $\max |z-(6+8i)|=12.$

Therefore, the correct option is: $\boxed{\text{B: }12}.$