I think you are looking for just 5 rounds of play as you mention each player having just one singles match. It is possible to arrange 5 rounds so that all possible opponents (combined over the doubles and singles matches) are seen exactly once each. Also a player's four doubles partners are all different.
(1 10 v 8 9) ( 7 4 v 2 3) (5 v 6)
(2 6 v 9 10) ( 8 5 v 3 4) (1 v 7)
(3 7 v 10 6) ( 9 1 v 4 5) (2 v 8)
(4 8 v 6 7) (10 2 v 5 1) (3 v 9)
(5 9 v 7 8) ( 6 3 v 1 2) (4 v 10)
This schedule on the Wiseman website (http://www.jdawiseman.com/papers/tournaments/individual-pairs/ip-asym_10.html) which is almost balanced for partners, may also interest you.
Ian.
9 rounds is a much harder problem and the singles matches can never be balanced - most players will have 2 singles matches, but 2 players will only have one. I can extend the idea from the schedule above to 10 rounds and get the following:
(9 8 v 6 1) (7 4 v 3 2) (5 v X)
(X 9 v 7 2) (8 5 v 4 3) (1 v 6)
(6 X v 8 3) (9 1 v 5 4) (2 v 7)
(7 6 v 9 4) (X 2 v 1 5) (3 v 8)
(8 7 v X 5) (6 3 v 2 1) (4 v 9)
(6 9 v 3 X) (1 4 v 8 2) (5 v 7)
(7 X v 4 6) (2 5 v 9 3) (1 v 8)
(8 6 v 5 7) (3 1 v X 4) (2 v 9)
(9 7 v 1 8) (4 2 v 6 5) (3 v X)
(X 8 v 2 9) (5 3 v 7 1) (4 v 6)
where player 10 is represented as "X". But like the 11 round tournament (see Wiseman link) a few pairs of players oppose just once while another few pairs oppose three times. If you leave out the 10th round above, I think you may have a workable solution, at least all the partnerships will be different, so one of your criteria above is met.