Experimenting with a CAS suggests an induction. In order to handle the induction, we need to consider the forms of the numbers involved. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ alternates $1$ and $0$ bits. The map $2n + 1 \to n - 2^{f(n)}$ drops the rightmost bit (which is $1$) and clears the next least significant bit. The map $2n \to n$ drops the rightmost bit (which is $0$). And the map $2n \to 2n - 2^{f(n)}$ moves the least significant bit right one place. So the numbers which occur in the evaluation tree of $\frac{4^m-1}{3}$ are of the form $2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k}$ where $i \le j - 2$ and $k \ge 0$; and (when we get down to one bit) the form $2^i$.
Starting with the simplest case, when $i > 0$, $a(2^i) = pa(2^{i-1}) + qa(2^{i-1})$ so by induction $a(2^i) = (p+q)^i$.
For the more general case, let $A(i,j,k) = a(2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k})$ subject to the aforementioned constraints on $i,j,k$.
For $i > 0$, we have an even argument and use the second case of the recursion:
\begin{eqnarray*}
A(i,j,k) &=& a(2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& pa(2^{i-1} + 2^{j-1} + 2^{j+1} + \cdots + 2^{j+2k-1}) + qa(2^{i-1} + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& pA(i-1, j-1, k) + qA(i-1, j, k)
\end{eqnarray*}
Then it's a trivial proof by induction that $$A(i,j,k) = \sum_{u=0}^i \binom{i}{u} p^u q^{i-u} A(0, j-u, k)$$
For $i = 0$, we have an odd argument and use the third case of the recursion:
\begin{eqnarray*}
A(0,j,k) &=& a(1 + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& a(2^{j+1} + \cdots + 2^{j+1+2(k-1)}) \\
&=& \begin{cases}
a(0) & \textrm{if } k=0 \\
a(2^{j+1}) & \textrm{if } k=1 \\
A(j+1, j+3, k-2) & \textrm{otherwise}
\end{cases} \\
&=& \begin{cases}
1 & \textrm{if } k=0 \\
(p+q)^{j+1} & \textrm{if } k=1 \\
\sum_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, k-2) & \textrm{otherwise}
\end{cases}
\end{eqnarray*}
Further CAS experimentation suggests that the theorem we need to prove is $$A(0, j, 2v-1) = A(0, j, 2v) = \left(p\frac{q^v-1}{q-1} + q^v\right)^{j+1} A(0, 2, 2v-2)$$
The first of those equalities is easy: $$A(0,j,2) = \sum_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, 0) = \sum_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} = (p+q)^{j+1}$$ so $A(0,j,1) = A(0,j,2)$ and since the only occurrence of $k$ in the third case is in the parameter $k-2$ the rest follows by induction on $v$.
The second equality is the interesting one. Again, by induction on $v$:
\begin{eqnarray*}
A(0,j,2v) &=& \sum_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, 2v-2) \\
&=& \sum_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} \left(p\frac{q^{v-1}-1}{q-1} + q^{v-1}\right)^{j+4-u} A(0, 2, 2v-4) \\
&=& \left[ \sum_{u=0}^{j+1} \binom{j+1}{u} p^u \left(pq\frac{q^{v-1}-1}{q-1} + q^v\right)^{j+1-u} \right] \color{Blue}{\left(p\frac{q^{v-1}-1}{q-1} + q^{v-1}\right)^3 A(0, 2, 2v-4)} \\
&=& \left( p + pq\frac{q^{v-1}-1}{q-1} + q^v \right)^{j+1} \color{Blue}{A(0, 2, 2v-2)} \\
&=& \left(p\frac{q^v-1}{q-1} + q^v\right)^{j+1} A(0, 2, 2v-2)
\end{eqnarray*}
as desired.
The answer to the original question now drops out: $a\left(\tfrac{4^m-1}{3}\right) = A(0, 2, m-2)$, but $A(0, 2, k)$ has the form of a cube times $A(0, 2, k-2)$ so by induction it's always a cube.
The conjectured formula can be proved by induction on $\mathrm{wt}(n)$.
For $\mathrm{wt}(n)=0$, we have $n=0$ and the conjectured formula trivially holds.
Now, for a positive integer $\ell$, suppose that the conjectured formula holds for all $\mathrm{wt}(n)<\ell$. Let us prove it for $\mathrm{wt}(n)=\ell$. So, let $n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_\ell+1}))\dots)$ with $t_j\geq 0$ and $n'=2^{t_2}(1+\dots(1+2^{t_\ell+1}))\dots)$, so that
\begin{split}
a(n) &= a(2^{t_1}(2n'+1)) = \sum_{m=0}^{t_1} \binom{t_1+1}{m} a(2^kn') \\
&=\sum_{m=0}^{t_1} \binom{t_1+1}{m} \sum_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} (1+\mathrm{wt}(j))^{m+t_{2}+1} \prod_{k=2}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t_{k+1}+1} \\
&=\sum_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} \left[(2+\mathrm{wt}(j))^{t_{1}+1} - (1+\mathrm{wt}(j))^{t_{1}+1}\right] \prod_{k=1}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t_{k+1}+1} \\
&=\sum_{j=0}^{2^{\ell-1}-1} \left[(-1)^{\ell-\mathrm{wt}(2j+1)}\prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j+1}{2^k}\rfloor))^{t_{k+1}+1} + (-1)^{\ell-\mathrm{wt}(2j)} \prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j}{2^k}\rfloor))^{t_{k+1}+1}\right] \\
&=\sum_{j'=0}^{2^\ell-1} (-1)^{\ell-\mathrm{wt}(j')} \prod_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j'}{2^k}\rfloor))^{t_{k+1}+1}, \\
\end{split}
where $j'$ corresponds to values $2j$ and $2j+1$ (in other words, $j=\lfloor j'/2\rfloor$). QED
Best Answer
As proved in this answer, if we represent $n$ as $n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_m+1}))\dots)$ with $t_j\geq 0$, then \begin{split} a(n) = P(\ell)^{t_1}\prod_{j=1}^{\ell-1} P(\ell-j)^{t_{2j}+t_{2j+1}+1}, \end{split} where $\ell:=\left\lfloor\frac{m+1}2\right\rfloor$ and $$P(k) := q^k+p\frac{q^k-1}{q-1}.$$ Notice that this does not depend on $t_m$ when $m$ is even.
Grouping the summands in $s(n)$ by the number of unit bits (very much like in this other answer), for $n\geq 1$ we have \begin{split} s(n) &= \sum_{m=1}^n\ \sum_{t_1+t_2+\dots+t_{m}=n-m}\ P(\lfloor(m+1)/2\rfloor)^{t_1} \prod_{j=1}^{\lfloor(m-1)/2\rfloor} P(\lfloor(m+1)/2\rfloor-j)^{t_{2j}+t_{2j+1}+1}\\ &= \sum_{m=1}^n\ [x^{n-m}]\ \frac1{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}\cdot\frac{1+\frac{(-1)^m-1}2x}{1-x}\\ &= [x^n]\ \sum_{m=1}^\infty\ \frac{x^m}{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}\cdot\frac{1+\frac{(-1)^m-1}2x}{1-x}. \end{split} That is, the generating function for $s(n)$ is $$\sum_{n\geq 0} s(n)x^n = 1+\frac1{1-x}\sum_{m=1}^\infty\ \frac{x^m(1+\frac{(-1)^m-1}2x)}{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}.$$
Combining terms for $m=2k-1$ and $m=2k$, we can rewrite it as $$\sum_{n\geq 0} s(n)x^n = 1+\frac{1}{1-x}\sum_{k=1}^\infty\ \frac{x^{2k-1}}{1-P(k)x}\cdot\prod_{j=1}^{k-1} \frac{P(j)}{(1-P(j)x)^2},$$ which is quite close to the conjectured formula.