It sounds like the system is using PWM dimming at a slow rate (1Hz, 50:50 duty cycle, and that this is fast enough that the decay in an incandescent bulb is not very noticeable (thermal time constants are long). If the car has a 12V electrical system, then the fact you measure 9V at this bulb is odd... Perhaps the ratio is 75:25?
You could try adding a (large) capacitor in parallel with the LED to create a lowpass filter with the R. The problem is that when the supply is off, all the LED current must be sourced by the capacitor. This will require a big capacitor. From a quick LTSPICE simulation, to bias a 2V LED 20mA average current from a 75:25 duty cycle 12V supply, this takes a 330 ohm series resistor and a shunt capacitor on the order of 100,000uF (in parallel with the LED). It's a little better with a blocking diode in series with the resistor, but not much. There is still a variation in the LED current of about 20%.
Perhaps you can trace this back and figure out what is providing the PWM to the light bulb and run the LED circuit from 12V with something else switching it on. For example, with a few transistors, you can build a nearly constant current source for the LED and turn it on with the PWM'd signal filtered with a series R and smaller C. I can post a diagram of this approach if you are interested.
Here's the circuit with no added capacitor, a 12V supply switched at 75% duty cycle, and about 20mA through a Vf=2V LED with R=360 ohm. The LED current just shuts off when the supply is off.
With a 100,000uF cap you can still see droop in the current, but the average is better. This would probably still have noticeable dimming. Note the current scale is different. Partly the current drops here because the LED is effectively regulating the voltage on vled when the supply is on. So the current drops rapidly when the supply is off.
Based on that observation, you really want an RC filter first and then the LED and its series resistor. This has much less ripple, and the cap can be a little smaller. Here it's 33,000uF.
It has been posted elsewhere, but there appear to be several reasons why LEDs can glow when the light is switched off. The underlying reason, of course, is that LEDs require very little power to illuminate. There is a video of someone standing under some transmissions lines and getting a LED to glow from the induced current.
The switch is not completely switching off; it is a dimmer or some other device that leaks some current when it is supposed to be off.
The switch is installed on the neutral wire. In this case, it isn't even necessary to have any leakage to earth. The cabling and circuitry acts as a capacitor that can have sufficient capacitance to allow a small current to flow.
There isn't a neutral wire. Though most countries have a neutral wire at near ground, there are some exceptions. Some old installations in Belgium and France have two hot wires. Hence (2) applies always.
If you have a dual-pole setup there are two live wires, one off at one end and one off at the other end. If the wire is long enough a current can be induced in the overlapping section of wire.
Best Answer
Your original schematic isn't clear but I think you're trying to use the N-Channel MOSFET as a switch. Since I don't know the current required by your LED strip you should spec one out yourself. This video gives a great introduction to MOSFETs. Below is an explanation for each of the components used.
R1 - current limiting resistor to limit the gate current to reduce voltage oscillations
R2 - pull down resistor to avoid spurious activations by creating a defined voltage level at all times
R3 - load resistor that represents the position of your LED strip in the whole schematic.
simulate this circuit – Schematic created using CircuitLab
As others have mentioned, it may be easier to just power the LED strip from a source that's controlled by the ignition. If it's a 12V supply you'll have to use a voltage regulator like the LM7805 to reduce the voltage. It's only rated for about 1A though. If your strip needs more consider a buck converter.