To traverse all the edges of a cuboid, each only once, requires the traverse of three diagonals and I could not immediately find such a path with all three on the same pair of faces with the same dimensions.  But even if such a path exists, it would cross on one of the faces which the teaser does not allow.

So, after a lot of sketching of possible paths, I believe that when all the diagonals are face diagonals, two of the three must lie on opposite cuboid faces with the third on another face. When one of the diagonals is a space diagonal, the two face diagonals do not need to be on opposite faces but there is always a path of this kind. (I have written a program which verifies these observations by generating all 1,421,716 different paths).

This means that two different path arrangements need to be considered:

In the left hand arrangement, the third diagonal is on another face, which means that the shortest integer length path traversing all edges of a cuboid with edges \((a, b, c)\) has a length of \[4(a+b+c)+2\sqrt{a^2+b^2}+\sqrt{a^2+c^2}\]where \((a,b)\) and \((a, c)\) are the non-hypotenuse sides of Pythagorean triangles.  In the right hand arrangement, the third diagonal is a space diagonal of the cuboid, giving the length as:\[4(a+b+c)+2\sqrt{a^2+b^2}+\sqrt{a^2+b^2+c^2}\].  In this case the third diagonal has to be what is known as a Pythagorean quadruple.