That such a decomposition doesn't exist can be shown by looking at the interior solid angles at the vertices.
The cube has eight interior solid angles of $\pi/2$ each.
Each of the six tetrahedra has one interior solid angle of $\pi/2$ and three that can be computed as shown here:
$$\Omega=\phi_{ab}+\phi_{bc}+\phi_{ac}-\pi=\pi/2 + \arccos\frac{1}{\sqrt{3}}+ \arccos\frac{1}{\sqrt{3}}-\pi=2\arccos\frac{1}{\sqrt{3}}-\pi/2\;.$$
Thus, the sum of the interior solid angles of the cube is $4\pi$, and the sum of the interior solid angles of the tetrahedra is
$$6\left(\pi/2+3\left(2\arccos\frac{1}{\sqrt{3}}-\pi/2\right)\right)=6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)\;.$$
This is not equal to $4\pi$, since $\arccos\frac{1}{\sqrt{3}}$ is not a rational multiple of $\pi$. (Here's an elementary proof of that fact.) In fact it is almost $5\pi$.
[Edit: I just realized in the shower that all the following analysis of potential inner vertices is actually not necessary. No matter what happens at the inner vertices, we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle $\pi/2$ are not enough to do it, so we have to use at least one of the others, but then we can't make the sum come out to $4\pi$, a rational multiple of $\pi$.]
This establishes that you can't fill the cube with the six tetrahedra by letting all their vertices coincide with the cube's vertices. It seems geometrically obvious that you can't do it with any of the vertices inside the cube, either, but this, too, can be proved rigorously using the solid angles. If there were a vertex inside the cube, the entire solid angle of $4\pi$ around that inner vertex would have to be filled. The candidates for filling it are a face of a tetrahedron, which subtends a solid angle of $2\pi$, an edge of a tetrahedron, which subtends a solid angle of twice the dihedral angle of the intersecting planes, i.e. either $\pi$ or $2\arccos\frac{1}{\sqrt{3}}$, or an interior solid angle of a vertex of a tetrahedron. Thus, adding up all the solid angles at the eight outer vertices and at $v$ inner vertices, the following equation would have to have solutions with non-negative integer values of $j$, $k$, $l$, $m$, $n$ and $v$:
$$6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)+j(2\pi)+k(\pi)+l\left(2\arccos\frac{1}{\sqrt{3}}\right)+m(\pi/2)+n\left(\arccos\frac{1}{\sqrt{3}}-\pi/2\right)$$
$$=4\pi+v(4\pi)\;.$$
This is impossible, since the coefficient in front of $\arccos\frac{1}{\sqrt{3}}$ is non-zero, and thus the equation would imply that this is a rational multiple of $\pi$, which it isn't (see above).
P.S.: I just realized I forgot to mention an essential part of the proof:
$$\cos 2\phi = 2 \cos^2 \phi - 1\;,$$
$$\arccos x = \frac{1}{2}\arccos(2x^2-1)\;,$$
$$\arccos \frac{1}{\sqrt{3}}=\frac{1}{2}\arccos(-\frac{1}{3})\;,$$
so $\arccos \frac{1}{\sqrt{3}}$ is rational iff $\arccos(-\frac{1}{3})$ is rational; then we can apply the theorem I linked to above.
Best Answer
The key insight is to first determine the minimum distance between the body diagonal and the face diagonal. Without loss of generality let the cube have unit side length and take the body diagonal to be the segment joining $(0,0,0)$ to $(1,1,1)$; then there are six face diagonals that do not intersect the body diagonal. These can be partitioned into two groups of three such that each group lies in a plane, and form the sides of equilateral triangle of side length $\sqrt{2}$ perpendicular to the body diagonal, which passes through the centers of these triangles.
Consequently, the minimum distance between the body diagonal and any such face diagonal is simply $$\frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}.$$
Since the tetrahedron is regular, with two vertices on the body diagonal and two on the face diagonal, this means distance between two non-adjacent edges of the tetrahedron is $1/\sqrt{6}$. If the side length of the tetrahedron is $2s$, then this implies the distance between $(s, -\frac{1}{2\sqrt{6}}, 0)$ and $(0, \frac{1}{2 \sqrt{6}}, s)$ is $2s$; i.e., we require $$s^2 + \frac{1}{6} + s^2 = 4s^2,$$ or $$s = \frac{1}{2\sqrt{3}}.$$
Therefore, the tetrahedron's volume is $$V = \frac{(2s)^3}{6\sqrt{2}} = \frac{1}{18\sqrt{6}},$$ and the volume for the original cube of side length $a$ is $$V(a) = \frac{a^3}{18 \sqrt{6}}.$$
We could also have found this by noting that the circumscribed cube to the tetrahedron has edge length $1/\sqrt{6}$, thus the edge of the tetrahedron, being also the face diagonal of the circumscribed cube, has length $\sqrt{2}$ times this. And the circumscribed cube's volume is just $\frac{1}{6 \sqrt{6}}$, which is $3$ times the volume of the inscribed tetrahedron, since the four congruent tetrahedra inside the cube but outside the regular tetrahedron each has volume $1/6$ that of the cube. So the desired tetrahedron's volume is again $1/(18 \sqrt{6})$.
For your understanding and enjoyment, please see the animation below, which illustrates the six tetrahedra that can be thus formed with a single body diagonal. A representative set of coordinates for a single tetrahedron, up to symmetry, is $$\left\{ \left(\frac{6 - \sqrt{6}}{12}, \frac{6 + \sqrt{6}}{12}, 0 \right), \left(\frac{6 + \sqrt{6}}{12}, \frac{6 - \sqrt{6}}{12}, 0 \right), \left(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right), \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \right\}.$$