It depends on how you define information, but I'd argue that it doesn't.
In particular, those digits are not random variables which are free to take on any value; they're fixed by the definition of sqrt(2). In terms of Shannon information, the information is a measure of uncertainty in a message. If our protocol is that I'll send you digits of sqrt(2), then there's no information being transferred at all: since you could have worked them out for yourself. Alternatively, if our protocol is that I'll send you digits with uniform probability, then sending sqrt(2) would be "infinite information", but only because we're having to narrow it down out of infinitely many possibilities. I don't think this is the most useful definition though, since we can arrange for any amount of information we like: if the protocol is I will either send "5" or sqrt(2), then it contains 1 bit; and so on.
Alternatively we could use algorithmic information theory, where the information content of a message is the length of the smallest tape for a universal turing machine which outputs that message. sqrt(2) can be calculated by a very small program (cranking out digits forever), so it contains very little information. Yet even here, since it's constant, we could define our turing machine such that it emits sqrt(2) when given an empty tape, and hence it again contains no information.
Let S be an encoding of the works of Shakespeare in binary. A normal number contains S in its binary expansion. We can do this for all information. In this sense is it correct to say that a normal number contains infinite information in it?
Again, depends on your definition of information. For shannon information, you would need to define a protocol. If the protocol is "I will send you the entire works of Shakespeare" then I would be sending you no information. Likewise, if the protocol is to send you S, then S contains no information.
