Warning: long post!
(EDIT: I've added another post correcting some errors in this one--please read both before responding!)
After thinking this over some more, I decided to bite the bullet and actually look at the stress-energy tensor of the pipe in the CoM rest frame and in the boosted frame.
Once again I found that Greg Egan has done most of the hard work, at his page on http://www.gregegan.net/SCIENCE/Rings/Rings.html" . I'm not going to go to the level of detail he does, but I recommend looking at his work as a check on what I'm going to say below about which components of the stress-energy tensor are nonzero.
Egan treats the case of a thin disk or hoop, but that will work fine for dealing with a thin slice of our pipe that's at a particular longitudinal coordinate. I'm going to use z instead of x for the longitudinal direction (the direction in which we're going to boost the frame), since the coordinates we'll end up using will be cylindrical, with r being the radial coordinate and \theta being the angular coordinate. All we'll have to do is add one term to Egan's stress-energy tensor, to deal with stresses in the z-direction (which he doesn't include).
Egan's web page actually expresses the stress-energy tensor in a frame which is "co-rotating" with the disk or hoop, but he gives enough information to make it easy to re-express it in the non-rotating frame which is at rest with respect to the pipe's CoM. When we do that, we find that the only nonzero components are (I'm not writing down the detailed expressions that Egan does; we're only interested here in which components are nonzero, not what their specific values are):
T_{tt} = E
-- energy density
T_{t \theta} = \Phi
-- angular momentum
T_{\theta \theta} = p_{\theta}
T_{rr} = p_r
T_{zz} = p_z
-- pressure in each of the coordinate directions.
The absence of any other nonzero components in the pipe's CoM frame corresponds to our assumption that there is no torsion in that frame. (I should note, here, that until I took this detailed look I hadn't really thought specifically about which stress-energy tensor components would indicate torsion; some of the discussion in this thread has indicated that torsion might have a timelike component--i.e., one of its indices would be t--but that is *not* correct, as we will see.)
Now, what happens when we boost this tensor in the z-direction? Boosting a tensor is kind of like boosting a vector squared; each component gets two coefficients from the boost matrix. Since we're boosting along the z-direction, our boost matrix, which I'll call L, has the following nonzero components:
L^{t}_{t'} = L^{z}_{z'} = \gamma
L^{t}_{z'} = L^{z}_{t'} = - \beta \gamma
L^{r}_{r'} = L^{\theta}_{\theta'} = 1
We then transform the tensor components using the boost matrix as follows: for a generic tensor component with indices a', b' in the boosted frame, transformation matrix L, and indices m, n in the original frame, we have:
T_{a' b'} = L^{m}_{a'} L^{n}_{b'} T_{mn}
Note that we are using the Einstein summation convention, so the right-hand side of the above is a sum over all m, n. So the components of our transformed tensor will be (writing only the non-zero components of each sum):
T_{t' t'} = L^{t}_{t'}^2 T_{tt} + L^{z}_{t'}^2 T_{zz}
T_{t' z'} = L^{t}_{t'} L^{t}_{z'} T_{tt} + L^{z}_{t'} L^{z}_{z'} T_{zz}
T_{z' z'} = L^{t}_{z'}^2 T_{tt} + L^{z}_{z'}^2 T_{zz}
T_{t' \theta'} = L^{t}_{t'} L^{\theta}_{\theta'} T_{t \theta}
T_{r' r'} = L^{r}_{r'}^2 T_{rr}
T_{\theta' \theta'} = L^{\theta}_{\theta'}^2 T_{\theta \theta}
Substituting, we obtain:
T_{t' t'} = \gamma^2 \left( E + \beta^2 p_z \right)
T_{t' z'} = - \beta \gamma^2 \left( E + p_z \right)
T_{z' z'} = \gamma^2 \left( \beta^2 E + p_z \right)
T_{t' \theta'} = \gamma \Phi
T_{\theta' \theta'} = p_{\theta}
T_{r' r'} = p_r
Notice that the only other nonzero term that appears is the t-z term, which corresponds to the pipe's momentum in the z-direction in the boosted frame. It is negative because in this frame the pipe is moving backwards (i.e., in the negative z' direction). As I noted above, this term does *not* represent torsion; it's just linear momentum.
Notice again that there are *no* other nonzero terms; i.e., *there is no torsion*! Torsion (i.e., shear stress) would look like a term with two different spatial indices, such as r-z, r-\theta, or z-\theta. There are no such terms in the boosted tensor above.
What does this mean? It means that, although the pipe "looks twisted" in the boosted frame (in the sense that a line marked on the pipe, which is parallel to the z-axis in the CoM frame, appears twisted around the pipe in the boosted frame), there is *no* torsional stress in the pipe corresponding to the apparent "twist". So the twist is, in that sense, an "illusion".
(EDIT: No it isn't--I missed a nonzero term in the boosted tensor above. See my next post.)