The $p_k$ and $q_k$ always appear together so let us for the moment write $a_k= p_kq_k$ (I will come back to the condition later).

Then your system can be rewritten as $a_{i+2} = 3a_{i+1} - 2a_i$ for $i$ a positive integer.
Now if you pick any two starting values $0 \lt a_1 \lt a_2$ you directly get from these equations a recursively definied sequence of positive integers $a_i$.

Now, you want in addition that $a_i = p_i q_i$ with $p_i,q_i \ge 3$. To get this notice that if $d$ divides both $a_{i+1}$ and $a_i$ then it also divides $a_{i+2}$. So pick your $a_1$ and $a_2$ such that the have a greatest common divisor that is the product of two positive integers at least $3$, say take $a_1=9$ and $a_2=18$. Then you can always factor $a_i$ as $a_i=p_i q_i$ with the condition you want.

**Added** in view of comments by OP, based mainly on comments by Emil Jeřábek and Barry Cipra (answer now in CW mode):

If one wishes to further analyse the possible solutions, one can notice that the above mentioned recurrence can be solved explicitly and the solutions are of the form $u2^i + v$.

In particular, one can note that successive terms of the sequence $a_i$ are relatively co-prime if and only if $v$ is odd and co-prime to $u$.

By a result of Selfridge it is know that for $v=1$ and $u=78557$ the sequence $u2^i + v$ never takes a prime value, and indeed the terms are always divisible by one of $3,5,7,13,19,37,73$.

Thus the terms of this sequence $78557 \cdot 2^n + 1$, with $i \ge 1$, always admits a decomposition as requested by OP, as a product of two numbers at least $3$.

I guess, but did not check, the proof of Selfridge's result is based on covering congruences, a method alluded to by ARupinski. That is one covers the integers by a finite set of congruence classes and then checks the different finitely many cases individually.