The answer is no, $\sum \epsilon_n a_n$ need not almost surely converge. For instance, with $a_n = \frac{(-1)^n}{\sqrt{n}}$, the random series $\sum \epsilon_n a_n$ converges with probability zero.

This follows from a very precise theorem of Rademacher-Paley-Zygmund, which says that the following are equivalent for $\{a_n\}$ any sequence of real numbers:

- The probability that $\sum \epsilon_n a_n$ converges is $1$.
- The probability that $\sum \epsilon_n a_n$ is bounded is $1$.
- $\sum a_n^2 < \infty$.

(Of course, to conclude that the series converges with probability $0$ in the alternative case, we appeal to Kolmogorov's $0,1$-Law.) I found this theorem stated as Theorem 10 in the notes "Almost sure limit sets of random series" by Pete L. Clark.