Andy,

I think the answer is no. There are 2 reasons. One is that on each day after the first there are always at least 2 pairs of people together who were together on at least one of the previous days. There just are not enough days to get all of the pairs. The second reason is that I used a computer to search through all of the possible schedules, and the “most balanced” one it found does not have the property you want. I’ve pasted it below anyway. The positions in each sequence are the players, and the numbers indicate the threesomes. You’ll notice that players 1 and 6 are never together.

Day 0 : (0, 0, 0, 1, 1, 1)
Day 1 : (0, 0, 1, 0, 1, 1)
Day 2 : (0, 1, 0, 1, 0, 1)
Day 3 : (0, 1, 1, 0, 0, 1)
Day 4 : (0, 0, 1, 1, 0, 1)
Day 5 : (0, 1, 0, 0, 1, 1)

As an aside, it does seem to be possible to get every pair together once if 7 rounds are played.

Sorry if this came too late to be useful.
—Victoria