I think the answer is - it depends.
Can your helicopter descend vertically without power at a steady speed?
Is that speed slow enough that it can be exchanged for rotor speed by increasing collective pitch just before touchdown?
If not - higher weight? smaller rotor? - then some forward speed is inevitable, to give more lift at lower collective pitch, so that the vertical speed is slow enough that it can be pulled away on touchdown.
In the video it looks like - from those heights/speeds and in that helicopter - forward speed is required. It never looks like a stable state of descent is reached in the time available (although the ground is out of shot), and so there's a juggling of rotor speed, forward and vertical speed throughout the descent.
A lot of relevant issues seem to be be raised in the text with
Height?velocity diagram - Wikipedia, the free encyclopedia
In steady vertical descent at speed v_d, potential energy is being exchanged for kinetic energy of the air beaten down through the rotor disk at v_f, and a little into turbulent motion too. From energy conservation M.g.v_d ~ A.rho.v_f.v_f.v_f/2, and from momentum/Newton-II M.g ~ A.rho.v_f.v_f. This gives v_f scaling with the square root of M/(A.rho), making it easier in dense air, with low weight and a big rotor, and gives a simplistic vertical speed of v_d ~ v_f/2. That's not comfortably distant from a vortex ring state - as v_f ~ v_d, rather than v_f >> v_d.
For the R44, A~75 sq-m, M~1000kg, so v_f ~ 10 m/s, v_d~5 m/s: I'd say that 5 m/s makes sense from the video.