If I counted correctly and didn't miss any, there are
arrangements. There is no particular method to this - just going through all possible combinations of rectangles, and then trying all arrangements.
An alternative method which does not rely on going through quite so many cases is as follows:
There are 12 internal edge segments, and we can remove some to form the rectangle partition. Not every combination of removed segments produces a valid partition however. It is invalid when at one or more of the four internal intersect points there are three removed segments, or if there are two removed segments at right angles. All other cases will result in a valid partition.
As far as I can see, enumerating just the valid combinations of those intersection points still involves a bit of case work. You could split it into cases depending on which sides of the central square are removed:
1. No sides removed. There are 3 possibilities for each intersection point, so $3^4=81$ valid rectangular partitions with a central 1x1 square. After applying Burnside's lemma to reduce by symmetry, there are $15$ left.
2. One side removed. There are 2 possibilities at the points on either side of the removed segment, and 3 at the other points, for $2^2 3^2=36$ possibilities. Reducing this due to the mirror symmetry there are $21$ unique ones.
3. Two opposite sides removed. There are 2 possibilities at each point, so $2^4=16$ arrangements but after symmetry reduction only $7$ unique ones remain.
4. Two adjacent sides removed. There are $3*2^2*1 = 12$ possibilities, which reduces to 7 after symmetry reduction.
5. Three sides removed. There are $2^2*1^2 = 4$ possibilities, which reduces to 3 after symmetry reduction.
6. Four sides removed. There is only $1^4 = 1$ possibility.
Adding these together, we get $15+21+7+7+3+1 = 54$ rectangular partitions.