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.
This answer requires some spatial examples, so I think it will be better if I generalize the case of a simple cube, which can be applied to all possible situations.
As you said, the total area covered by the paint in a cube with lenght $a$ is equal to:
$$
S_{\mathrm{ext}}=6\cdot a^2
$$
But then the division into $27$ cubes with lenght $1/3$ of the original gives the new surface area, greater than the original because new slices prodce more internal surface; so now the total surface is expressed by:
$$
S_{\mathrm{tot}}=27\cdot 6\cdot (a/3)^2=\frac{27\cdot 6}{9}\cdot a^2=3\cdot 6\cdot a^2=3S_{\mathrm{ext}}
$$
Now think about a Rubik's Cube, with only the cubes facing the outside are painted: since you know that the total amount of paint occupies the external surface $S_{\mathrm{ext}}$, then to know the surface not covered in paint you have to simply subtract to the total sliced surface $S_{\mathrm{tot}}$ the external one:
$$
S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=3S_{\mathrm{ext}}-S_{\mathrm{ext}}=2S_{\mathrm{ext}}=12\cdot a^2
$$
Now you have basically derived the sum of the internal surface of every little cube, because the sum of the external is nothing more than the total covered in paint, or $S_{\mathrm{ext}}$ (with the Rubik's Cube case, the only cubes coloured).
EDIT: If the number of slices is not defined, it is generalized to a single parameter $n\in \mathbb{N}$, then the new surface tends to be greater as $n$ gets larger.
Let's assume that one cut is repeated on every edge of the cube, so if an edge is divided in three (with two slices), then the number of cubes created is expressed as follows:
$$
N_{\mathrm{cubes}}=(n+1)^3
$$
So if I "slice" a cube for every edge once ($n=1$), I get $N_{\mathrm{cubes}}=(2)^3=8$ little cubes.
The surface of each sub-cube is expressed by:
$$
S_{\mathrm{sub-cb}}=6\cdot\left(\frac{a}{n+1}\right)^2
$$
Then the total surface is expressed as th single one multiplied by the total number of sub-cubes created $N_{\mathrm{cubes}}$:
$$
S_{\mathrm{tot}}=(n+1)^3\cdot 6\cdot\left(\frac{a}{n+1}\right)^2=(n+1)\cdot 6\cdot a^2=(n+1)\cdot S_{\mathrm{ext}}
$$
Now, applying the same deduction as before, the surface not covered with paint is given by a simple subtraction between the total surface and the ponly painted one, which is the external one:
$$
S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=(n+1)\cdot S_{\mathrm{ext}}-S_{\mathrm{ext}}=n\cdot S_{\mathrm{ext}}
$$
Now remember that the parameter $n$ is the number of slices, and not the sub-cubes created; as your previous example, the slices were $2$, and so this number appeared inside the final result.
Best Answer
The short answer is yes, although to be rigorous we should use the word "measure" instead of volume. For our purposes, measure is the $n$-dimensional analogue of volume; in one dimension, it is length, in two dimensions it is area, in three dimensions it is volume, and in general $n$-dimensions it is the "generalization" of these concepts.
Measure has the intuitive property of countable additivity: if we take an at most countably infinite collection of pairwise disjoint sets $\{A_i\}$, then the measure of $\bigcup_{i = 1}^\infty A_i$ should be the sum of the measure of each $A_i$ separately. Basically, the volume of two non-overlapping cubes is just the sum of the volume of each cube separately. Then allow for "two" to be replaced with "a countably infinite number of" cubes, and that is countable additivity. It also has the intuitive property that if $A \subset B$, $m(A) \subset m(B)$, where $m(A)$ is the measure of $A$.
The 4D hypercube has the property that at each slice perpendicular to the $w$-axis, the result is a 3D cube. This is the higher-dimensional analog to the fact that if we slice a 3D cube perpendicular to the $z$-axis, we get a 2D square.
Now, let $C$ denote the 4D hypercube. We write this as $$ C = [0, 1] \times [0, 1] \times [0, 1] \times [0, 1]. $$ Letting the first coordinate be $w$, note that $C$ is the disjoint union $$ \bigcup_{w \in [0, 1]} (\{w\} \times [0, 1] \times [0, 1] \times [0, 1]). $$ This is, in fact, an uncountable union of a disjoint collection of sets, each of which is a 3D cube with 3D measure = 1. Let $A = \{a_1, a_2, \ldots\}$ be a countable subset of $[0, 1]$. Then by countable additivity, and the fact that the measure of a subset is smaller than the measure of the set containing it, we have the following: \begin{align*} m(C) &= m\left(\bigcup_{w \in [0, 1]} (\text{3D-cube})\right) \\ &\geq m\left(\bigcup_{w \in A} (\text{3D-cube})\right) \\ &= \sum_{w \in A} m((\text{3D-cube})) \\ &= \sum_{w \in A} 1 = \infty. \end{align*} Thus, the 3D measure $m$ of the 4D cube is infinite.