Following Arturo's interpretation (2), you can still get a handle on the order type. Specifically, it is $\omega+\mathbb{Q}\times \omega$. The first $\omega$ is the strings consisting of all 0's. Any string containing at least one 1 can be considered the binary expansion of a dyadic rational in $(0,1)$ ending in 1 followed by some number of 0's. The dyadic rationals in $(0,1)$ are order-isomorphic to $\mathbb{Q}$ by the usual ordering (which agrees with the lexicographic ordering after you trim the trailing 0's). The number of trailing 0's gives you the remaining factor of $\omega$.
Arturo's example gives an infinite decreasing sequence in the "$\mathbb{Q}$" part, so you won't get an ordinal order type, but the rationals are considered one of the "basic" order types, so it's not so bad.
EDIT: As Marian points out, the math is right, but the typography is wrong. It is, in fact, $\omega+\omega\times\mathbb{Q}$. See i.e. http://en.wikipedia.org/wiki/Ordinal_multiplication, which in turn cites Jech.