The 'at least once' property that you are looking for is not mathematically possible. The best that can be done is to have three pairs of players who do not play together, while other pairs meet once or twice. In the schedule below the former are (1 6), (2 4) & (3 5).
(3 4 1) (5 6 2)
(1 2 5) (4 3 6)
(6 5 4) (2 1 3)