This will work perfectly fine. As long as you don't make the MOSFET's body diode conduct with a reverse voltage greater than a few hundred mV, there won't be any difference between a simple NMOS transistor and a proper transmission gate / analog switch.
In other words: Just keep the peak-to-peak amplitude of the signal at the BC549's emitter below 600mV or so (300mV in either direction).
If you want the circuit to function with slightly larger signals as well, you could do something like this:
simulate this circuit – Schematic created using CircuitLab
This makes the circuit's biasing more robust as the common-emitter amplifier's gain is now 1 at DC (instead of 10), and it also decreases the voltage across the MOSFET, giving you more headroom before the body diode begins to conduct.
If you can fit an additional MOSFET, you can also use this configuration to avoid the body diode problem entirely:
simulate this circuit
This switch configuration only requires the gate voltage to get significantly higher than the maximum signal voltage in order to turn it on, and lower than (or equal to) the signal voltage to turn it off. For example, if you have a signal that swings between 0V and 2V at the BJT's emitter, a gate voltage swing of 0V / 12V would be more than sufficient to cleanly turn the two-MOSFET switch on or off.
Note, however, that you might need a stronger gate driving signal to avoid charge coupling from the floating source node into the gate. You could, for example, drive the gates hard to GND / VCC with a SPDT switch. If you do this, you should still add a pull-down resistor (i.e. 100k) to prevent the gates from floating during the switch transition. Additionally, you'll need to use a break-before-make switch.
It's also quite important for this configuration that the position of the capacitor is swapped as shown, as this prevents negative voltages from developing at the two-MOSFET switch's terminals. Otherwise you'd need a negative gate voltage to turn it off reliably.