Probably this is wildly known, but the closest I found was a surface with additional restrictive conditions.

A perfect cuboid is a cuboid having integer side lengths, integer face diagonals and an integer space diagonal leading to positive integer solutions of:

$$a^2+b^2=s_1^2\qquad \qquad \rm(1)$$ $$a^2+c^2=s_2^2\qquad \qquad \rm(2)$$ $$b^2+c^2=s_3^2\qquad \qquad \rm(3)$$ $$a^2+b^2+c^2=s_1^2+c^2 =s_4^2\qquad \qquad \rm(4)$$

wlog one can work with rationals and dividing by $c^2$ one can assume $c=1$ making (2) and (3) $a^2+1=s_2^2$ and $b^2+1=s_3^2$ and they can be parametrized by the substitutions $a=\dfrac{u^2-1}{2u}$ and $b=\dfrac{v^2-1}{2v}$

This leaves only (1) and (4). $s_1^2+1=s_4^2$ leads to the next parametrization $s_4=\dfrac{s^2+1}{s^2-1}$ and $s_1=\dfrac{2s}{s^2-1}$.

This makes (1) and (4) equal and the denominator vanish for $u=0$,$v=0$ or $s= \pm 1$.

So the surface comes from the numerator of (1):

$u^{4} v^{2} s^{4} + u^{2} v^{4} s^{4} - 2 u^{4} v^{2} s^{2} - 2 u^{2} v^{4} s^{2} - 4 u^{2} v^{2} s^{4} + u^{4} v^{2} + u^{2} v^{4} - 8 u^{2} v^{2} s^{2} + u^{2} s^{4} + v^{2} s^{4} - 4 u^{2} v^{2} - 2 u^{2} s^{2} - 2 v^{2} s^{2} + u^{2} + v^{2} = 0$

The trivial rational points contain $0$ and $\pm 1$.

(A) Does the surface contain all perfect cuboids?

(B) Is there a reason/heuristic to believe the surface might have non trivial rational points?

**EDIT** Per John's answer some congruences were found on the homogenized surface

$u^{4} v^{2} s^{4} + u^{2} v^{4} s^{4} - 2 u^{4} v^{2} s^{2} h^{2} - 2 u^{2} v^{4} s^{2} h^{2} - 4 u^{2} v^{2} s^{4} h^{2} + u^{4} v^{2} h^{4} + u^{2} v^{4} h^{4} - 8 u^{2} v^{2} s^{2} h^{4} + u^{2} s^{4} h^{4} + v^{2} s^{4} h^{4} - 4 u^{2} v^{2} h^{6} - 2 u^{2} s^{2} h^{6} - 2 v^{2} s^{2} h^{6} + u^{2} h^{8} + v^{2} h^{8}$

$$ v^{2} \cdot u^{2} \cdot s^{4} \cdot (u^{2} + v^{2}) \equiv 0 \mod {h}$$ $$ (s - h)^{2} \cdot (s + h)^{2} \cdot v^{2} \cdot h^{4} \equiv 0 \mod u$$ $$ (s - h)^{2} \cdot (s + h)^{2} \cdot u^{2} \cdot h^{4} \equiv 0 \mod v$$ $$ h^{4} \cdot (u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} h^{2} + u^{2} h^{4} + v^{2} h^{4}) \equiv 0 \mod s$$

The substitution $ s = u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} + u^{2} + v^{2}$ leads to a factor:

$$ u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} + u^{2} + v^{2} = 0$$

This is genus 1 curve with known trivial points, and it is of rank 0 (if I have done the computations right).