Capacitor value for rectified magneto PSU
I'm not familiar with motorcycle generators, but I'm going to assume the generator runs at engine RPM, which can be anything from 500 to 10000 rpm... Well, your vintage bike probably doesn't go that high, still this is a quite wide range, corresponding to 8-100 Hz. That's approximative.

Now, the capacitor's role is to smooth voltage at low rpm so your light keeps steady.

At 500rpm (8 Hz) the capacitor is charged 8 or 16 times per second, depending on what waveform your magneto generates, of which I have no idea. I'll just go with 10 times per second. 

The 9W LED will draw 1A if powered by 9V, a nice round number.

The capacitor's equation is i = C dv/dt

Let's go with: choose a capacitor value such that the i=1A current draw will cause a voltage drop of dv=1V during dt=0.1 second. This results in a 100000µF capacitor, which is impractically large. Try a 10000µF 40V cap first, that will cost you a few bucks.

Now, the voltage is a problem. Your generator will have an open circuit voltage which can be quite high when the light is off. Also, it depends on RPM, and since the wave shape won't be a sine, a multimeter in AC mode won't be helpful.

For example, a "6V 3W" bicycle hub dynamo will happily put out 50V unloaded when speeding downhill. Open circuit voltage is proportional to rpm, so make sure to put the switch BEFORE the capacitor. If you put the switch after the cap, then the cap will charge to the rpm-dependent open circuit voltage, which will most likely exceed 30V, and this will exceed your LEDs's maximum voltage when you close the switch.

Now, you have another problem, which is these crude generators usually work with incandescent lights. So, if you rev up the engine, the generator's power increases, voltage increases a bit, but the light will draw more current, which compensates.

However, a LED with a switchmode electronic driver is different: as input voltage increases, it draws LESS current. In fact it will draw a constant power. If it is a 9W LED, if you give it 9V it will draw 1A, but if you give it 18V it will draw 0.5A.

So, what's going to happen is that when you rev up the engine, voltage will increase, causing the LED to draw less current, which will cause the voltage to increase some more, until the LED blows.

So, first we will determine if the boost converter will be necessary. 

You need to connect the rectifier and capacitor to the generator, start the bike, and measure the DC voltage on the cap when the engine is idling. If it is 6V then you'll need the booster. If it is more than 9V, connect the LED. 

Now, most likely it will blink because the generator doesn't have enough power at low rpm to supply 9W, and the only fix for this is a LED with a driver which is smart enough to lower its intensity when voltage is too low. Maybe yours does? You should check.

If your LED electronic driver is dumb, it will turn on at full power as soon as it gets 9V, which will cause the voltage to go down, and then it will turn off. A boost converter will not fix this, as the problem in this case is that the generator does not provide enough power at low RPM. The fix is to have a LED which draws less power when input voltage is low.

Next, SLOWLY increase the rpm, and monitor voltage on the cap. Write it down. Don't exceed the LED's maximum.

Most likely you will find that at some rpm which isn't that high, you'll get more than 30V, thus your problem will be to avoid overvoltage, and you will need a shunt regulator to fix this.

So, get a diode bridge, a cap and a multimeter, and when you have measured numbers, you'll have your solution.