Not really. Let's see.

Assume, without loss of generality, that all guests are real numbers within [0, 1], just to apply the diagonalization properly.

If the (countably) infinite hotel is full, one can use diagonalization to create a new guest and put it in the hotel.

But... How many guests can be created, given a specific guest list? The answer is lim (n -> oo) 9^n = 9^N, since, for each digit of the diagonal number, there are 10 - 1 = 9 choices of a substitute digit. A set of cardinality 9^N is already known to be uncountable, because 9^N > 2^N, the cardinality of base-2 numbers from 0 to 1.

In short: you can use diagonalization to add ever more guests to the Hilbert Hotel, but only a countable set; the vast majority of the waiting list won't ever enter.