There are 10 combinations of suits HH, DD, CC, SS, HD, HC, HS, DC, DS, SC (Note CS, for example, doesn't count as a different combination to SC for these purposes. When the order does matter it is called a permutation).
So the Guesser asks:
Is the card a heart?
Is the card a diamond?
Is the card a club?
Is the card a spade?
If the answer is yes to any of these then the Guesser guesses that suit. If not then,
Is the card either a heart or a diamond?
Is the card either a heart or a club?
Is the card either a heart or a spade?
Is the card either a diamond or a club?
Is the card either a diamond or a spade?
Is the card either a spade or a club?
The Tester must answer yes to at least one of these options, namely the one which lists the suits of both cards.
The Tester must answer yes at least once to these 6 questions and must answer no at least three times, but may answer yes or no to four of the six.
If the Tester answers yes only once then you know the suits of both cards and may guess.
If the Tester answers yes twice then one suit will be present in both of the combinations, guess that suit.
If the Tester answers yes five times out of six then the cards are of the two suits different to the two suits which get the no answer.
If the Tester answers yes four times the two no answers will cover three suits, guess the fourth suit.
If the Tester answers yes three times then there are 6 combinations of pairs. Two combinations have a suit present in all three pairs, guess that suit. Two combinations will have two suits which are present twice and two suits which are present once each, guess either of the suits present twice. Two combinations have three suits present twice and one suit not present.
So this is where I'm stuck. From here you have a 2/3 chance to guess correctly, but I can't think of a way to get a definitive answer if the Tester answers yes three times. At least assuming that the Tester will be consistent when asked the same questions again. Otherwise if you had two rounds of three yes answers which had both of the combinations that have a suit absent then you know both absent suits which tells you which two are present of course.
I will probably come back to this when my head is clear. UGH! I hate to leave it when so many of the possible scenarios are solved!