All such schedules will have a minimum of 6 pairs of competitors who never meet.  This is an unavoidable problem that is 
discussed here.
So I think the following is about the best you can do:
 (10  9 v  4  2)  ( 3 12 v  6  8)  (11  7 v  5  1)                                        
 ( 6 10 v  5  3)  ( 4 12 v  7  9)  (11  8 v  1  2)                                        
 ( 7  6 v  1  4)  ( 5 12 v  8 10)  (11  9 v  2  3)                                        
 ( 8  7 v  2  5)  ( 1 12 v  9  6)  (11 10 v  3  4)                                        
 ( 9  8 v  3  1)  ( 2 12 v 10  7)  (11  6 v  4  5)  
Players play in partnership with 5 of the other players exactly once, and in opposition with a different 5 players exactly twice.  In other words no pairs of competitors are meeting three times.