It's possible to arrange that all pairs of couples meet together at the same house either two or three times each.
(12 7 9 3) ( 4 10 6 11) ( 5 8 1 2)
( 8 6 4 12) (10 2 9 5) ( 7 11 3 1)
( 3 9 5 6) ( 2 7 1 4) (12 11 8 10)
( 8 9 7 11) (10 3 2 4) ( 1 12 5 6)
(11 2 6 9) ( 7 5 4 12) ( 3 1 8 10)
( 6 1 10 7) ( 9 4 3 8) (11 12 2 5)
( 9 4 11 1) ( 5 8 10 7) ( 6 3 12 2)
( 4 5 11 3) ( 2 6 7 8) ( 1 10 12 9)
I think that may solve your problem. If you had 11 months then it would be possible to do better and have all pairs of couples meet exactly three times each.