Yes I do believe it would work well. A mount could be made with a quick attach plate. A hydraulic motor of the correct CC to equal the speed that the blade should turn. The 422 and the 425's Hydraulic PTO puts out 8 gallon per minute. I had to figure this out when I made my side mower out of a 28 inch snapper mower deck and blade. It has worked very well and has been used extensively mowing pond banks, bushes, and ditches.
The Hesko Storm Front has a 24" x 5" fan; front and side discharge chutes. Selection of front or side discharge can be made from the operator's seat. Fan speed is approximately 2200 RPM, producing up to 4200 CFM
1 gallon [US, liquid] = 231 cubic inchs
8 gallon would = about 1848 cubic inches
Theoretically 1848 cubic inches would be pumped in one minute. You need 2200 rpm.
1848 ci / 2200 rpm = .84 cubic inch motor or any where close should be alright.
Some people might say to multiply this ans. by .85 to account for a motor efficiency of 85 %. But in my case for the side mower I didn't and it seemed to have turned out perfect.
Be sure to go over these figures to be sure this would be right.
It might help if you look over this tread.
http://www.tractorbynet.com/forums/power-trac/64961-side-mower.html
Pumping: 1 Hp = 1 GPM x 1500 Psi or say 2 GPM @ 1500 Psi = 2 Hp