don't take my word for it - draw up the circuit in LTspice
I will probably do that later in Tina-TI; I installed that recently on Thor's recommendation to compare to LTSpice.
But for the moment let's take as a predicate that the phase splitter with capacitor on the drain works as described, and the cold side connection has a LPF behavior while the hot side does not.
Just as background to make sure everyone reading can follow along, the standard no filtered output to a diff amp; for easy math let's say the signal is normalized to an amplitude of 1 on each side at every frequency of interest, i.e. hot side signal amplitude is 1, cold side signal amplitude is -1 (same absolute value, but opposite polarity).
A difference amplifier takes the difference, i.e. subtraction, of the cold signal from the hot, so the output amplitude is:
1 - (-1) = 2
For completeness I show the case of LPF in cold side and hot side. For ease of math assume the filter response is -6dB at 2kHz, -12dB at 4kHz.
At low frequencies the amplitude will still be 1, so the output of the difference amplifier will be 1 - (-1) =2.
at 2kHz the output amplitude is 0.5 and -0.5, so the output of the difference amplifier will be 0.5 - (-0.5) =1
at 4kHz the output amplitude is 0.25 and -0.25, so the output of the difference amplifier will be 0.25 - (-0.25) = 0.5.
In dB that makes the 2kHz response 20*log(1/2) = -6dB
and at 4kHz 20*log(0.5/2) = -12dB
Now let us add a low pass filter in only the cold side, again for ease of math assume essentially flat at low frequencies, -6dB at 2kHz, and -12dB at 4kHz (6dB/octave 1st order filter). If the beginning amplitude is 1, then the -6dB amplitude will be 0.5 and the -12dB amplitude will be 0.25. The filter is in the return/cold side, so our convention is negative amplitude.
At low frequencies it will be the same as the unfiltered case, 1 - (-1) =2
at 2kHz it will be 1 - (-0.5) = 1.5
at 4kHz it will be 1 - (-0.25) = 1.25
So in dB at 2kHz, where the output of the cold side signal is -6dB, the output of the difference amplifier will be 20*log(1.5/2) = -2.5dB
At 4kHz where the output of the cold side signal is -12dB, the output of the difference amplifier is 20*log*1.25/2) = -4dB
The difference between linear and logarithmic shows up there, the output of the difference amplifier is not actually down 3dB at the 6dB point of the LPF, and it isn't quite 6dB down at the 12dB point of the LPF, but the slope is obviously reduced compared to the case
So it would (somehow) end up as 3dB/octave (as opposed to 6dB/oct, 1st-order filter slope)? How?
Hopefully that explanation is easy enough to follow.