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.
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.
Best Answer
Quite similarly to my answer to the previous question, we have that for $n=2^tk$ with odd $k$, $$ a(n)=\sum_{i=0}^t \binom{t}{i}p^{t-i}q^i a(2^i(k-1)+1). $$
It further follows that for $n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_s+1}))\dots)$ with $t_j\geq 0$, we have \begin{split} a(n) &= \sum_{i_1=0}^{t_1} \binom{t_1}{i_1} p^{t_1-i_1}q^{i_1} \sum_{i_2=0}^{t_2+t_3+1+i_1} \binom{t_2+t_3+1+i_1}{i_2}p^{t_2+t_3+1+i_1-i_2}q^{i_2} \sum_{i_3=0}^{t_4+t_5+1+i_2} \\ &\qquad\dots \sum_{i_\ell=0}^{t_{2\ell-2}+t_{2\ell-1}+1+i_{\ell-1}} \binom{t_{2\ell-2}+t_{2\ell-1}+1+i_{\ell-1}}{i_\ell}p^{t_{2\ell-2}+t_{2\ell-1}+1-i_\ell}q^{i_\ell} \\ &=\prod_{j=0}^{\ell-1} \bigg(q^{\ell-j}+p\frac{q^{\ell-j}-1}{q-1}\bigg)^{t_{2j}+t_{2j+1}+1}, \end{split} where we conveniently define $\ell:=\left\lfloor\frac{s+1}2\right\rfloor$ and $t_0:=-1$.
Now, for $n=\frac{2^{kn}-1}{2^k-1}$ we have $s=n$, $\ell=\left\lfloor\frac{n+1}2\right\rfloor$, $t_1=0$ and $t_j=k-1$ for all $j\in\{2,3,\dots,s\}$, implying that $$a(\tfrac{2^{kn}-1}{2^k-1}) = \prod_{j=1}^{\lfloor (n-1)/2\rfloor} \bigg(q^{\lfloor (n+1)/2\rfloor-j}+p\frac{q^{\lfloor (n+1)/2\rfloor-j}-1}{q-1}\bigg)^{2k-1}.$$ In particular, setting $k=1$, we get $$a(2^{n}-1) = \prod_{j=1}^{\lfloor (n-1)/2\rfloor} \bigg(q^{\lfloor (n+1)/2\rfloor-j}+p\frac{q^{\lfloor (n+1)/2\rfloor-j}-1}{q-1}\bigg),$$ and thus $$a(\tfrac{2^{kn}-1}{2^k-1}) = a(2^{n}-1)^{2k-1}.$$