I ran a small computer program to find lists of nine digits that sum to 45 and have a sum of squares of 285. One of these is of course 9,8,7,6,5,4,3,2,1 but there could be others.
And indeed there are. Here are the 25 possibilities my program found:
9,8,7,6,5,4,3,2,1 * <- normal set of digits
It may not be immediately obvious that this shows there are false Sudoku solutions that pass the limited validity check in question. Take a valid Sudoku solution, and replace all the 1s by the letter A, 2s by B, etc. Now choose one of the 24 abnormal sets of digits from one of the rows above. Do another substitution, changing the letters A-I by that abnormal set of digits in any order. Each house, row, and column will have one complete set of those digits and so will sum to 45 and have a squared sum of 285.
Edit: As Neil W pointed out, if you also add the check that the sum of the cubes adds to 2025, then there are still two false sets of digits, which I have marked with an asterisk in the list above.
Edit2: I also did a search for all sets of nine digits with sum 45 and product 9!. This leads to only one false set of digits, namely 9,9,7,5,4,4,4,2,1.