Diode voltage drop is only .7 volts which will be pretty much insignificant to the formula power(P) = voltage(E) times current(I). P=E^2/R
I don’t believe that running 2 diodes in parallel is a great idea since the slightest change in characteristics is going to cause one diode to do most of the work.
These days higher amp diodes are not that expensive. I see some 15 A for under 2 dollars.
Since you don’t know the full characteristics of the element then the easiest way to get the actual current would be to set up the circuit and measure the current with a meter. Otherwise just over design and go for worst case and then some.
Re: voltage drop...
The Schottky diode has a relatively small voltage drop, usually between 0.15 to 0.45 V.