This is slightly easier to express in terms of probability. A string with r streaks has r-1 changes, that is, r-1 locations where the ith symbol is different from the (i+1)st symbol. In a random string, these changes occur independently with probability 2/3, so the count is a Binomial(n-1,2/3) variable. The probability mass function for a binomial variable is well-known, P(k successes out of n) = (n choose k) p^k (1-p)^(n-k).
(# strings with r changes) / (3^n total strings) = (n-1 choose r-1) (2/3)^(r-1) (1/3)^(n-r).
After some algebra (be careful that the denominators on the right have product 3^(n-1)), we get
(# strings with r changes) = 3 * (n-1 choose r-1) 2^(r-1).
You can think of the right hand side as saying that a string with r streaks can be encoded by the first symbol, the locations of the r-1 subsequent streaks out of the n-1 locations after the first, and for each new streak we have a choice of two symbols different from the previous symbol. We could use this to count the strings without probability, but I think it's better to recognize the connection with the binomial variable.
