No. No current flow at start-up. The diode has current flow at turn off to help discharge the cap.When the switch closes, the diode allows the capacitor to charge instantly,
Isn't the diode pointing the wrong way for current to flow from the capacitor to the + power lead? My understanding of the diode symbol is, the arrow with the line blocking it indicates which direction classical current (i.e. positively charged, before people discovered electrons were negatively charged) is blocked. In this case you have the diode blocking classical current from flowing towards the + power lead.
The supply will never be higher than 9V in this application.You could add a moderate value fixed resistor across the FET to give a fast discharge path - eg. 1K.
If the supply is less than 15V or so, you probably don't then need R2, so it stays quite simple.
That's the exact circuit I was imagining in my earlier answer.