What hexahedra have faces with areas of exactly 1, 2, 3, 4, 5, and 6 units?

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.