I replace $\alpha$ by $x$ below.
Restricting ourselves to positive numbers is useless, since the property considered does not depend on $a_0(x) = \lfloor x \rfloor$. Therefore, $B$ is the union of all images of $B \cap [0,1[$ by integer translations.
Now, call $G$ the Gauss-Kuzmin map from $[0,1[ \setminus \mathbb{Q}$ to itself: $G(x) = 1/x-\lfloor 1/x \rfloor$.
This map preserves the probability measure $\mu$ with density $x \mapsto (\ln 2)^{-1}(1+x)^{-1}$ on $[0,1[ \setminus \mathbb{Q}$ and is ergodic.
The partial quotients $(a_n(x))_{n \ge 1}$ of $\alpha$ are the integer parts of the iterates $(G^n(x))_{n \ge 1}$. Hence $(a_n)_{n \ge 1}$ and $(q_n)_{n \ge 1}$ are Borel functions, so $B$ is a Borel set.
One checks that $\ln(a_1+1)$ is integrable with regard to $\mu$. Birkhoff theorem applies, so when $n \to +\infty$
$$\frac{\ln q_n}{n} \le \frac1n\sum_{k=1}^n \ln(a_k+1) \to C:= \int\ln(a_1+1)\mathrm{d}\mu \quad \mu \text{-a.s..}$$
On the other hand, $(q_n)_{n \ge 1}$ is bounded below by Fibonacci sequence. Therefore, by comparison, the series $\sum_n \frac{\ln q_{n+1}}{q_n} $ converges $\mu$-almost surely, and also almost surely for the Lebesgue measure.