I have to be honest and say that I don't fully understand your query, with a SELECT in there like that, so it that may be OK. But I think you want to use LIKE '%google%'. In any case, I probably would have done this:
Code:
UPDATE user LEFT JOIN usertextfield ON (user.userid = usertextfield.userid)
SET usergroupid = 3
WHERE signature LIKE '%google%'
AND posts = 0
AND lastvisit = 1342328885