No, it will turn the LED OFF when it is running.
To invert it's operation you could remove the LED alltogether, connect the emitter to 0V, increase the size of R4 to about 10K and connect the gate of your MOSFET to the collector of Q3.
When the transistor turns on, it will pull the bottom of R4 to near 0V. When the fan is working it will be about 12V.
I don't know how efficient it would be.
Edit: I can try and work out worst case losses in the circuit, but it's a bit of a guess. Somebody correct my numbers please.
I'm assuming Vsupply = 12V and the duty cycle of the fan is 50%. I'm assuming that the circuit is used for one fan only and that a neglegable amount is lost through the capacitor.
the fan is connecting the junction between R1 and R2 to ground at a duty cucly of 50%. Therefore the power lost (Through R1) is:
P = V^2 / R * DutyCycle
P = 12*12 / 10,000 * 50%
P = 144 / 10,000 * 0.5
P = 0.0072A = 7.2mA
In adition to this, an amount is lost through R1, R2, D2 (The zener) and Q3 while the fan is in a fail condition (Neglegable while the fan is working okay. (We'll ignore D2 and Q3):
P = V^2 / R
P = 12*12 / 20,000
P = 144 / 20,000
P = 0.0072A = 7.2mA
Finally, in the inverting circuit I suggested above, we have a loss theough R4. However this loss is neglegable while the fan is working fine:
P = V^2 / R
P = 12*12 / 10,000
P = 144 / 10,000
P = 0.0144A = 14.4mA
So in total, aproximatly:
While fan is okay: 7.2mA
While the fan is in a fault condition: 28.8mA
There will also be a small loss from charging and discharging C1 in both conditions.
I guess you could increase R1 to 100K but you'd have to try it.