What hexahedra have faces with areas of exactly 1, 2, 3, 4, 5, and 6 units? I tried for a while, not  very hard, to construct a polyhedron with exactly six faces, whose areas were respectively 1, 2, 3, 4, 5, and 6 units. I  did not meet with any success. Still, it seems that it should exist, because the space of possibilities is so large and so weakly constrained. Perhaps you could make one by chopping off two of the vertices of a tetrahedron.
To be more specific, I do not care whether the hexahedron is regular or whether its faces are regular, or the same shape.  I would prefer that it be convex.
 A: Arrange six vectors with lengths $(1,2,3,4,5,6)$ head to tail so that they form a closed loop in $\mathbb{R}^3$.  As you say, there is a large space of possibilities here.  One method would be to first arrange them in a plane to form a loop, and then kinking a few into 3D.
You can form a planar loop by inscribing the chain in a large circle and shrinking the radius of the
circle until the chain closes to a loop.
One you have these six non-coplanar vectors, apply Minkowski's Theorem:

Theorem (Minkowski). Let $A_i$ be positive faces areas and $n_i$ distinct, noncoplanar unit face normals, $i=1,\ldots,n$. Then if $\sum_i A_in_i =0$, there is a closed polyhedron whose faces areas uniquely realize those areas and normals.

(See Chap. 7, p. 311: Aleksandr D. Alexandrov. Convex Polyhedra. Springer-Verlag, Berlin,
2005. Monographs in Mathematics. Translation of the 1950 Russian
edition by N. S. Dairbekov, S. S. Kutateladze, and A. B. Sossinsky.)

I wrote a note on this:
"Convex Polyhedra Realizing Given Face Areas,"
arXiv:1101.0823 [cs.DM], 4Jan11.  Here is a suggestive figure from my paper,
which hints at one method of arranging the vectors in space:
             
Some computational aspects of Minkowski's Theorem are discussed in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, p.340.
Of course you don't need that generality to solve this specific problem instance.
