A random Dirichlet series has its line of convergence as a natural boundary with probability $1$. The particular series coming from $\sqrt{2}$ that you are looking at is overwhelmingly *unlikely* to have an analytic continuation beyond $\sigma > 1$ (Correction: I was too hasty; the series is *likely* to have a meromorphic continuation to $\sigma > 1/2$. See the objection by David Speyer and my corrected answer.) The Law of Large Numbers does *not* show that almost all reals in $[0,1)$ correspond to Dirichlet series that converge for $\sigma > 1/2$. In fact, the set of reals for which the summatory function of the binary digits is $o(x)$ obviously is a null set. So the line of convergence is $\sigma = 1$ with probability $1$ and this is then a natural boundary with probability $1$. (Correction: In this situation with random choices from $0,1$, $\sigma = 1$ will be the line of convergence, but not a natural boundary, with probability $1$. See the objection by David Speyer and my corrected answer.)

You have been misled by the analogy with the Moebius function. The latter *oscillates* around zero, which your binary expansion coefficients do not.

By the way, the Law of Large Numbers is too weak to draw the kind of conclusions that you want. You need something stronger, for example the Law of the Iterated Logarithm.

I would expect that the Dirichlet series with coefficients $c_n = 2b_n - 1$ where $b_n$ is the n-th binary digit of $\sqrt{2}$ should converge in $\sigma > 1/2$ and have $\sigma = 1/2$ as a natural boundary.