I will assume that it will be OK if one family of 4 plays the other family of 4.
If not we will subtract that 1 situation.  It would be much better in this type
of problem if people would use definite wording like "there are exactly 2", or
"there are at least 2" instead of the ambiguous "there are 2", which could be
taken as meaning either.  [If you have 3 dollars, you have 2 dollars, right?]

Case 1:  The Gomez brothers are on the same team, and the Zhang brothers are on
the other team.  

Then we have 2 boys on each team, and two from the same family on each team, so
we can choose any 2 of the 4 girls to play on the Gomez brothers' team.  That's
4C2 = 6 ways to choose a team for case 1, because the remaining 2 girls will
play on the Zhang brothers' team, in only 1 way. 

For the other two cases, notice that:
One of the Gomez brothers is younger than his older brother.
One of the Zhang brothers is younger than his older brother.

Case 2:  The younger Gomez boy is on the same team with the younger Zhang boy,
and the older Gomez boy is on the other team with the older Zhang boy.

Then we can choose any 2 girls to play on the team with the two younger boys,
because any of the 4 girls is sister to one of those boys.  So that's 4C2 = 6
ways to choose a team for case 2, because the other 2 girls will play on the
older boy's team.  They will both be a sister to one of the boys, making 2 from
the same family on each team.  

Case 3:  The younger Gomez boy is on the same team with the older Zhang boy, and
the older Gomez boy is on the other team with the younger Zhang boy.

That's just like case 2.  So that's also 4C2 = 6 ways.

Answer: 6+6+6 = 18 ways.

Edwin