Will an inductor wound using wire of 0.6438 mm (22 AWG) be close enough to 0.508 mm (25 SWG) for a neglectable difference?
The inductor wound with 0.508 wire would have a length of 4.064 (perfectly wound).
The inductor wound with 0.643 wire would have a length of 5.15   (8 turn coil)

Difference in length is 1.09 mm.  

This is about a 27% difference in length.  Inductance therefore is going to be reduced by about 27%. Not a trvial amount. Increasing inductor length using same amount of turns (wider copper) results in reduced inductance (Inductance inversely proportional to Length). 

Inductance would equal  0.789 of the original designed inductance.   

One could try using 9 turns of the larger diameter wire, attempting to get inductance back to normal.
Inductance is proportional to the number of turns squared. 

\$(9^2/8^2)*0.789\$ = 0.99 of original designed inductance.  

Advanced users : I have kept it simple here.  Of course for 8 turns one uses 9 times the copper diameters for coil length.  And obviously the "perfect" winding with no space between turns. 

It doesn't often turn out this simple to change turns to accommodate change in copper width (coil length).  Just fortunate that one additional turn (in this case) can restore original inductance.