This doesn't come close to resolving the problem, but here's a small subproblem that might be useful to consider. For convenience, when I say "aperiodic" I mean "not eventually periodic."
Pick an arbitrary subset $T\subseteq\mathbb{N}$ and define
$$s_n=\left\{\begin{array}{ll}
n+1 & \text{ if }n\in T,\\
n & \text{ if }n\notin T.
\end{array}\right.$$
This sequence is non-decreasing and so we can recover a valid sequence of non-negative integers $a_n=s_n-s_{n-1}$. For general $T$ (i.e. $T$ that is not the union of a finite set and finitely many arithmetic sequences), $a_n$ is aperiodic. Now
\begin{align*}
S&=\sum_{n=0}^\infty(\tfrac 13)^n(\tfrac 49)^{s_n}\\
&=\sum_{n=0}^\infty(\tfrac 13)^n(\tfrac 49)^{n}+\sum_{n\in T}(\tfrac 13)^n\left((\tfrac 49)^{n+1}-(\tfrac 49)^{n}\right)\\
&=\frac{27}{23}-\frac59\sum_{n\in T}(\tfrac {4}{27})^n,
\end{align*}
so $S$ is rational if and only if $\sum_{n\in T}(\tfrac {4}{27})^n$ is rational. In other words, if the original question has an affirmative answer, then the following must hold: For any base-$\frac{27}{4}$ expansion using only $0$ and $1$, the value is rational if and only if the expansion is eventually periodic.
More generally, while the original problem has two scaling factors, we can restrict to a problem with a single scaling factor by starting with a periodic sequence $a_n$ (which will allow us to combine the two scaling factors into a single product factor) and then introducing small aperiodic perturbations. Only a very narrow class of aperiodic sequences arise in this way, but if the claim is true in general, it certainly must be true for this narrow class.
Now to illustrate some of the difficulties that arise even for the restricted problem. Schmidt 1980 Theorem 2.4 proves that if the greedy $\beta$-expansion of every rational number in $[0,1)$ is eventually periodic, then $\beta$ is an algebraic integer. (In an earlier comment I cited a different paper, but that was a mistake: the result I cited there is only about purely periodic expansions.) In other words, for any non-integer rational number $\beta>1$, there exists a rational number that has an aperiodic base-$\beta$ expansion. So if we were to replace $\frac{27}{4}$ with rational $1<\beta<2$ (which ensures that the greedy base-$\beta$ expansion uses only coefficients in $\{0,1\}$), the bold statement above would be false.
There is of course a substantial difference between $1<\beta<2$ (when every real number has at least one base-$\beta$ expansion using only coefficients in $\{0,1\}$) and $\beta>2$ (when the set of real numbers that have a base-$\beta$ expansion using only coefficients in $\{0,1\}$ has measure zero). The greedy base-$\frac{27}{4}$ expansion of a rational number may use coefficients as large as $6$, so it's still possible that the statement in bold is true without contradicting Schmidt's theorem. But proving irrationality for even a single aperiodic expansion (let alone all of them) seems very difficult: for comparison, Terry Tao's blog post points out that the rationality of $\sum_{p\text{ prime}}\frac{1}{2^p-1}$ is unknown, and I would guess that $\sum_{p\text{ prime}}\frac{1}{(27/4)^p}$ is not any easier.
