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.
As achille hui pointed out in a comment, the relevant generalization here would be De Gua's theorem for three dimensions or the Conant–Beyer theorem for arbitrary dimensions. Colloquially speaking it states that the squared measure of your object equals the sum of the squared measures of its projection on a set of mutually orthogonal hyperplanes, in your case the coordinate axes.
So let's look at this in the dimensions we know. A segment from $(a,0)$ to $(0,b)$ has some length $d$ as measure. Projecting the segment onto the $x$ axis yields a segment of length $a$, projecting onto the $y$ axis yields a segment of length $b$. So taken together you have $d^2 = a^2 + b^2$ as expected.
In three dimensions, you have a triangle, and project that onto the coordinate planes. But here the area of the projection is a triangle, not a rectangle, so you don't simply multiply the intercepts, but divide by $2$ as the area of a right triangle is $\frac12 ab$. Thus the area is
$$s^2 =
\biggl(\frac{ab}2\biggr)^2 +
\biggl(\frac{ac}2\biggr)^2 +
\biggl(\frac{bc}2\biggr)^2 =
\frac{a^2b^2+a^2c^2+b^2c^2}{4}
$$
This differs from the $s^2=\frac14\left(a^2+b^2+c^2\right)$ you claimed here originally, but fits in with the $s^2=\frac14((ab)^2+(bc)^2+(ac)^2)$ you wrote in the duplicate post.
For four dimensions, you have to turn the volume of the cuboid into the volume of the simplex spanned by three of its edges. The factor to divide by here is $6$, so the formula for the volume will be
$$v^2 =
\biggl(\frac{abc}6\biggr)^2 +
\biggl(\frac{abd}6\biggr)^2 +
\biggl(\frac{acd}6\biggr)^2 +
\biggl(\frac{bcd}6\biggr)^2$$
So where you assumed the denominator to be $9$, it is in fact $6^2=36$. That $6$ there in the denominator is the $3!$ achille hui wrote in the comment, which is a more useful notation because it also explains how you get the corresponding factor for arbitrary dimensions. You might think of the area of the simplex as $\frac13$ times base plane times hright, where the base plane is again $\frac12$ times one length times the other. If you go to $d$ dimensions, you take all ways of combining $d-1$ of the $d$ edge lengths, multiply them and divide them by $(d-1)!$. The resulting numbers get squared and added.
If you want a general formula for arbitrary dimensions, you could write something like
$$v^2=\sum_{i=1}^d\left(\begin{array}{c}
\displaystyle\prod_{\substack{j=1\\j\neq i}}^{d}x_j
\\\hline(d-1)!\end{array}\right)^2$$
This is for the simplex spanned by $(x_1,0,\dots,0)$ through $(0,\dots,0,x_d)$.
Best Answer
Reposting my comment as an answer:
If I interpret your question as "is there a scissors congruence between the four dimensional polytopes $D^2$ and $A^2 \sqcup B^2 \sqcup C^2$", the answer is yes.
I can take every step of the proof at cut-the-knot and replace the equality between volumes by a scissors congruence. Roughly speaking, every step in the argument either replaces a triangle with base $b$ and height $h$ by a $b \times (h/2)$ rectangle, or replaces the square on the hypotenuse of a right triangle by the union of the two squares on the sides. (More precisely, it does these things and then does some algebra, which can be interpreted as taking the products and disjoint unions with some other polytopes.)
Both replacing a triangle with base $b$ and height $h$ by a $b \times (h/2)$ rectangle, and replacing the square on the hypotenuse of a right triangle by the union of the two squares on the sides, have decomposition proofs. Compose all of those decompositions and you'll get a proof of this result.