ANSWER: Obviously, if you believe FARADAY'S LAW, there is no emf around the loop. In principle this would not exclude a current flowing through R, if there were a battery at the other end to provide the voltage rise to go with its voltage drop. So the question really boils down to whether you can use an electric field as a battery. This sounds silly until you start asking where the charges go, at which point it gets confusing enough to raise doubts. So that's what this is really all about. Where do the charges go? Well, there are definitely surface charges on the wire (although none on the inside; it is, after all, a conductor, and inside). Positive surface charge will accumulate on the top of the loop and a corresponding amount of negative charge will go to the bottom, inside the capacitor. In fact, the entire potential drop between the capacitor plates will have to occur in the gaps between the wire and the plates, since the wire is an equipotential.1 So why don't the electrons on the bottom zip around through R to meet up with their positive mates on top and provide a nice perpetual motion machine? Because the same plate's electric field is holding them back! As we move outside the capacitor, "bulges" outward, giving a component parallel to the wire that pulls electrons back toward the plate on the bottom and pulls any positive charges back toward the plate on the top. When we get far enough away from the capacitor for to be negligible, the charges are happy to stay where they are. This question is potentially confusing only because it is so simple and we aren't used to thinking in these terms.
ANSWER: A truly rigourous calculation is daunting, but we can make a "spherical elephant" approximation that gives the correct qualitative behaviour without bogging down in details or being "glib". Model the tube as a stack of rings. (It isn't, of course, but the main effect does not depend on conduction along the tube axis .) Now model the magnet as a simple dipole. Let's assume the North pole is on top so that the magnetic field lines from the magnet point upward along the centre of the tube, as shown. They also, of course, spread out away from the axis, so that as the magnet moves downward, less magnetic flux links a given "ring" of the tube above the magnet. This generates an emf in that ring tending to generate an "upward" flux to replace what went away - i.e. a current into the page on the right and out of the page on the left. At the position of the ring, has an "outward" horizontal component, which gives the Lorentz force (on the induced current in the ring) a downward component all around the ring. NEWTON'S THIRD LAW demands an equal upward force on the magnet due to the current in the ring, but we have recently seen suspicious results from a naive application of said law to magnetic forces, so we should check its veracity with alternate arguments. One would be that the magnetization of the magnet is equivalent to a current ring that goes into the page on the right and out on the left to produce an upward - just like the induced current in the ring! Parallel currents attract, so the magnet is pulled upward by the induced current's field. OK, looks like Newton is safe for now. A ring below the magnet sees an increasing flux as the magnet approaches, and so "fights the change" with a current in the opposite sense: in on the left and out on the right. This repels the effective current ring of the magnet, so the rings below it also impede its descent. Are we done? Not quite. How come it still falls? If the tube were a perfect conductor, any induced emf would generate a current big enough to completely compensate any attempted change in flux, and a strong enough magnet would simply stop falling. It is only because copper has a finite conductivity that the ring ever comes out the bottom; the "drag" force increases as gravity accelerates the magnet, eventually reaching a terminal velocity determined by the magnet's strength to weight ratio and the conductivity of the tube.
There's no "correct answer" to this one!