Curious $q$-analogues Consider the Fibonacci polynomials
$$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j} $$
and the Lucas polynomials
$$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\frac{n}{n-j }\binom{n-j}{j} x^{n - 2j} .$$
Let $X$ be the multiplication operator $Xp(x)=xp(x)$ on the polynomials, $D_q$ the $q$-differentiation operator defined by $$D_q p(x)=\frac{p(qx)-p(x)}{qx-x}$$ and
$A=X+(q-1)D_q$.
Applying the operator $F_n(A)$ to the constant polynomial $1$ has the curious effect that all binomial coefficients $\binom{n}{j}$ are changed to $q$-binomial coefficients ${n\brack j}$ together with some $q$-power.  More specifically, we get 
$$F_n(A)1=F_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }
{q^{j+1\choose 2}}{n-j\brack j} x^{n - 2j} $$
and
$$
L_n(A)1=L_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }
{q^{j\choose 2}}{\frac{[n]}{[n-j]}}{n-j\brack j} x^{n - 2j} .$$  
Is this an isolated phenomenon or do there exist similar formulae for polynomials? In other words, are there polynomials $p_n(x)$  in one variable $x$  and operators $B(q)$  satisfying $B(1)=X$ such that $p_n(B(q))1$ is a simple or beautiful $q$-analogue of $p_n(x)$?  
A trivial example would be  $p_n(x)=x^n$ and $B(q)=\epsilon(q)X$ with $\epsilon(q)p(x)=p(qx)$.
Edit
Recently I noticed that the Rogers-Szegö polynomials
$$R_n(x,s)= \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j} $$
can also be represented in this way:
$$R_n(x,s)=(x+s+(q-1)xsD_q)^n1.$$
Since this is the most natural $q-$ analogue of the binomial theorem I would be astonished if nobody has as yet seen this fact.  The proof is the same as that for the well-known recursion $R_{n+1}(x,s)=xR_{n}(x,s)+(q^n-1)xs R_{n-1}(x,s).$
It follows immediately from the recurrence of the $q-$ binomial coefficients:
$(x+s+(q-1)xsD_q) \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j}$
$=\sum_{j = 0}^{n }{n\brack j} x^{j+1}s^{n-j}$
$+\sum_{j = 0}^{n }{n\brack {j}} x^{j}s^{n-j+1}$
$+\sum_{j = 0}^{n }(q^{j}-1){n\brack j} x^{j}s^{n-j+1}$
$= \sum_{j = 0}^{n+1 }({n\brack j}+{n\brack {j-1}}+(q^{j}-1) {n\brack {j}}) x^{j} s^{n-j+1}$
$=\sum_{j = 0}^{n +1}{{n+1}\brack j} x^{j}s^{n-j+1}.$
 A: This is not a definitive answer, since it doesn't contain any more examples, but more of an observation.
If you have a sequence $P_n(x)$ for polynomials defined recursively by the formula
$$P_n(x)=x^{a_1}P_{n-1}(x)+x^{a_2}P_{n-2}(x)+\cdots+x^{a_k}P_{n-k}(x),$$
and an operator $B(q)$ with $\lim_{q \to 1} B(q)=X$, we get:
$$P_n(B)1=B^{a_1}P_{n-1}(B)1+B^{a_2}P_{n-2}(B)1+\cdots+B^{a_k}P_{n-k}(B)1.$$
Denoting $P_n(B)1$ as $\hat P_n(x)$ and supposing that $B$ is linear (which it is, if it depends only on $X$ and $D_q$, like all your examples suggest), we get:
$$\lim_{q\to 1}\hat P_n(x)=\lim_{q\to 1}x^{a_1}\hat P_{n-1}(x)+x^{a_2}\hat P_{n-2}(x)+\cdots+x^{a_k}\hat P_{n-k}(x),$$
Thus, $\lim_{q\to 1}\hat P_n(x)$ satisfies the same recursion as $P_n(x)$ does and if the first $k$ terms are equal, then $\lim_{q\to 1}\hat P_n(x) = P_n(x)$ for all $n$. This is precisely the definition of a $q$-analog. Whether it will be interesting is another matter, but I think this must be decided case-by-case.   
I noticed that all of your examples have the form $B(q)=X+a(q-1)D_q$. This looks somewhat like a linear term of a series in $D_q$ of the form
 $$B(q)=X+a_1(q-1)D_q+a_2(q-1)^2D_q^2+\cdots,$$
 so maybe if you could find examples with $B$ not linear in $D_q$, it would be very interesting.
