The cutoff frequency f of this Sallen-Key 2nd-Order Low-Pass Filter is
f = 1 / 2(pi)RC
No, that's not quite true if we use the definition of the cut-off frequency being the point where the signal power at the output has halved (3 dB down). When both resistors and both capacitors are the same value, the filter has 6 dB attenuation at f = 1 / 2(pi)RC.
This is because like all other similar 2nd-order low-pass filters the transfer function magnitude at f = 1 / 2(pi)RC equals Q, the quality factor of the circuit: -
And if you made R1=R2=R and C1=C2=C, Q is 0.5
Hence the TF is 6 dB down at f = 1/2(pi)RC
For equivalent values of RC, do different combinations of R & C give
different signal integrity in practice.
Absolutely. If you are prepared to make the capacitor C1:C2 ratio 2:1 you get a Q of 0.7071 and although this doesn't sound much, you get a 3 dB attenuation at the cut-off frequency - this is called the butterworth maximally flat response. If you make the ratio bigger you get higher Q and a bigger resonant peak in the pass-band: -
So, if you are going to use a Sallen key filter but keep resistors and capacitors the same value you are missing a trick to get the best frequency response i.e. the flat butterworth response. In fact you might as well just cascade two RC filters and get rid of the op-amp entirely if you are not going to identify and use the benefits that a controlled peaking circuit can provide. OK using an op-amp gives you good buffering so it's not entirely wasteful!