You still have to solve 20 of them, no matter how many cubes you say you are going to do.
I heard a suggestion a while ago that was based on the WCA time limit for cubes. Remember how for multiple blindfold you get 15 minutes per cube for each cube up to 6, and 10 minutes for each after that? The idea was that you give a large time penalty, say about 10 or 20 minutes, for each cube that is unsolved at the end, and then check if the competitor is still under the time limit for the number of cubes that ended up solved. Presumably if they are not the attempt would count as a DNF or something.
The way this would work is - suppose you did 20 cubes out of 100. Well you didn't solve 80 of them, so you add on 80*20 minutes = 26 hr 40 minutes penalty. Now no matter what your time is this is well over the time limit for solving 20 cubes, so your attempt wouldn't count as a successful solve. That's why you can't just do a huge number of cubes. But if you look at Momonga's 17/18 in 2:10, that would be turned into a successful solve of 17 cubes, out of 18, in a time of 2:30 including the penalty - although he made an honest mistake he did solve 17 cubes correctly in under the time limit for solving 17 cubes, so in this system of measurement it would count as a record.
I don't do multiple blindfold, but I've done enough statistics to know that, even if someone has a constant rate of success no matter how many cubes they attempt, 17/18 actually represents a higher rate of success than 2/2. If you get a cube right 71% of the time (very low), you can get 2/2 half of the time, but the chances for 17/18 are practically infinitesimal... It seems odd to me to give 2/2 a higher ranking given that it is both more likely for a cuber to get it, and requires less skill.