The resistor is not a good idea. The current or 10 μA is the maximum current that the LED will pass at 5 V reverse. It will probably pass far less than that, and the resistor will no drop nearly that much voltage. I realise that the data sheet says that the reverse current is typically is 10 μA, but I simply don't believe it. It will be far less.
Any resistor that significantly reduces the voltage with the LED in reverse will make it just about invisible forward.
The LED will probably be fine at 9V. If you search this forum for transformerless power supplies, and running LEDs from mains, you'll see that many LEDs can stand really big reverse voltages.
What you want is a standard diode. A 1N4148 or 1N4001 can survive 50 V in reverse and will definitely prevent damage to the LED, and will only reduce the voltage by 0.7 V in the forward direction, so a small reduction in the series resistor will restore the current to what you want. You should maybe put a large value resistor, maybe 1 MΩ, in parallel with the LED to make sure that the reverse voltage is near zero even if the standard diode has a tiny bit of leakage current.