The answer to your question is: no

Its also easy to give a concrete counter-example:
Take the cube with vertices
$$
\left(\frac{273}{340},\,\frac{79}{68},\,\frac{13}{20}\right) ,
\left(\frac{407}{340},\,\frac{57}{68},\,\frac{27}{20}\right) ,
\left(\frac{239}{340},\,\frac{789}{884},\,\frac{337}{260}\right) ,
\left(\frac{249}{340},\,\frac{621}{884},\,\frac{217}{260}\right) ,
\left(\frac{263}{340},\,\frac{1195}{884},\,\frac{289}{260}\right) ,
\left(\frac{417}{340},\,\frac{573}{884},\,\frac{231}{260}\right) ,
\left(\frac{431}{340},\,\frac{1147}{884},\,\frac{303}{260}\right) ,
\left(\frac{441}{340},\,\frac{979}{884},\,\frac{183}{260}\right)
$$ inside a $0-2$-cube. Then the three volumes you ask for are $\frac{24421}{8840}\neq \frac{24563}{8840}\neq \frac{24491}{8840}$.
Here is a picture of the configuration:

The outer cube has volume 8 and the inner cube has volume $1/8$. As a sanity check, lets see if everything adds up:
$$\frac{24421}{8840} + \frac{24563}{8840} + \frac{24491}{8840} = \frac{14695}{1768}$$
and $$\frac{14695}{1768} + \frac{1}{8} -8 =\frac{193}{442}.$$

Since the edges of the outer cube are not coplanar with the edges, we have now double counted all the 12 tetrahedra, that arise from the fact that these pairs of edges are non-coplanar. Their volumes are
$$\frac{471}{8840} ,
\frac{11}{170} ,
\frac{59}{2210} ,
\frac{29}{680} ,
\frac{49}{1768} ,
\frac{29}{8840} ,
\frac{29}{680} ,
\frac{29}{8840} ,
\frac{59}{2210} ,
\frac{49}{1768} ,
\frac{11}{170} ,
\frac{471}{8840} $$ and sure enough the sum of these numbers is $\frac{193}{442}$.
Here is a picture of all those tetrahedra:

...and a picture of a single one:

By the way: those two cube are concentric, so your conjecture does not work even in the concentric situation.