Your question presupposes that collapse actually occurs. A third observer might view Bob and Alice to be entangled with the system. The combined state lives in a "larger" Hilbert space (I mean in the sense that it is a product space). Now, there are projections of this state into spaces where Alice particle 1 to have gone down path a, Bob particle 2 to have gone down path B, etc, and also vice versa. If you buy this (a lot of people don't), then there is no collapse from the third observer's point of view. Also, the choice of projection is largely a human construct (although motivated by things like pointer states), and so we don't need the theory tell us anything about human constructs.
Clarification:
Suppose we send two particles through a beam splitter (I don't really know why we need more than one particle, but ok). For a single particle, you would just have
$$\left|\Psi_1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left|\Psi_{1,left}\right\rangle + \left|\Psi_{1,right}\right\rangle \right)$$
But if we say that they are entangled, then we need something more general:
$$\left|\Psi \right\rangle = c_{ll} \left|\Psi_{1,left}\right\rangle \otimes \left|\Psi_{2,left}\right\rangle + c_{lr} \left|\Psi_{1,left}\right\rangle \otimes \left|\Psi_{2,right}\right\rangle + \mathrm{2\,more\,terms}$$
I don't think this represents "decoherence" in any case. This setup can still produce interference patterns. Decoherence is when such interferences are washed out by having too many degrees of freedom that are out of phase.
Now, suppose we have two detectors, called A and B, that detect particles. A can only detect particles that went left, and B can only detect particles that went right. When the two particles leave the splitter, the combined system is in the state
$$\left|\Psi\right\rangle \otimes \left|A_\mathrm{ready}\right\rangle \otimes \left|B_\mathrm{ready}\right\rangle$$
An external observer picks a time slice later, that, in her reference frame, she considers that $A$ and $B$ have had enough time to perform detection. Remember, quantum states are not reference frame invariant. In that frame, the state looks more like
$$c_{ll} \left|\Psi_{1,left}'\right\rangle \left|\Psi_{2,left}'\right\rangle \left|A_{click}\right\rangle \left|B_\mathrm{ready}\right\rangle + c_{lr} \left|\Psi_{1,left}'\right\rangle \left|\Psi_{2,right}'\right\rangle \left|A_{click}\right\rangle \left|B_\mathrm{click}\right\rangle + \dots$$
Now, this is a superposition of different possible outcomes. If you say that detector A registered a click, then you project onto $\left| A_\mathrm{click}\right\rangle$, and you are left with only the first two terms of the above state. From that point of view (human construct), B is in a superposition. But you can do that for detector B also, and find A to be in a superposition. The theory just predicts what the theory predicts. you are interpreting this projection business as collapse.
But to answer your question, can you detect this superposition. Well, in principle, if detectors A and B then go on to deflect a photon with some mirrors and things, you could recover an interference pattern when you do this over and over again. In practice, there are other parts of the system that we didn't write down, and those all contain their own messy phases, which never quite line up, which makes it impossible to detect. People have been trying with things like SQUIDs and the like, but it is difficult to prevent decoherence.
