It is enough with one continuous function. First, I'll give a simple example with one function which is discontinuous at one point. To do it, consider the function $$f:(0,\pi+1)\to(0,\pi+1)$$ with
$$
f(x) = \begin{cases}
x+1 &\text{if $x<\pi$,} \\
x-\pi &\text{if $x>\pi$,} \\
1 &\text{if $x=\pi$.}\\
\end{cases}
$$
Claim:
The sequence
$$1,f(1),f^2(1),\dots \tag{$*$}$$ is dense in $(0,\pi+1)$.
To verify the claim, it is enough to see that the image is dense in the interval $(0,1)$, and that is true because for every $n$, the number $\lceil n\pi\rceil-n\pi$ is in the image, and the sequence of multiples of $\pi$ modulo 1 is dense in $(0,1)$ due to $\pi$ being irrational.
Let $A$ denote the image of the sequence $(*)$.
Since $A$ is dense in $(0,\pi+1)$, we can find an homeomorphism $h:(0,\pi+1)\to\mathbb{R}$ with $h(A)=\mathbb{Q}$ (using that $\mathbb{R}$ is countable dense homogeneous, see for example this reference). We can also suppose $h(1)=0$ changing $h$ by $h-h(1)$ if necessary.
Then the function $F=hfh^{-1}$ does the trick, because $$F^n(0)=hf^nh^{-1}(0)=h(f^n(1)),$$ so $h(A)$, which is $\mathbb{Q}$, is the image of the sequence $0,F(0),F^2(0),\dots$
To prove that the problem can be solved with one continuous function, we can apply the same argument but taking instead of $f$ a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $0,g(0),g^2(0),\dots$ is dense in $\mathbb{R}$. As Martin M. W. noticed in his answer, those functions are known to exist (they are called transitive maps), this paper gives examples of them.
As was mentioned in the comments by pregunton, it is possible to do using two rational functions. I claim it is not possible using just one. As Fedor Petrov suggests in another comment, this is because rational functions of degree higher than $1$ are never going to be surjective, which can be shown with help of Hilbert's irreducibility theorem. Indeed, take a rational function $\frac{f(x)}{g(x)}$ with coprime polynomials $f$, $g$ of which at least one has degree greater than $1$. The polynomial $h(x,t)=tf(x)-g(x)\in\mathbb Q[x,t]$ is irreducible then, so by Hilbert's theorem there are infinitely many values $q\in\mathbb Q$ for which $h(x,q)\in\mathbb Q[x]$ is irreducible. For all but one of these $q$, $h(x, q)$ will have degree $\max(\deg f,\deg g)>1$, so irreducibility implies it has no rational roots. Hence $q$ is not in the image of $\frac{f(x)}{g(x)}$.
The only case remaining is that of $\deg f,\deg g\leq 1$. In this case either $\deg g=1$ and the rational function has a rational pole, so its iteration can't go over all rationals, or else it is affine of the form $ax+b$ and it's easy to see explicitly its iterations do not cover all rationals.
Best Answer
Yes. This answer is based on the answers to your previous question.
Start with a computable ergodic map $T$ (D. Thomine constructs an example here). For every basic open neighborhood $B_i$, the set $D_i = \bigcup_n T^{-n}(B_i)$ is a dense effectively open set, uniformly in $i$. So some computable real $p$ meets all the $D_i$, meaning $X = \{ T^n(p) : n \in \omega\}$ is dense.
There is a computable, order-preserving bijection $g: X \to \mathbb{Q}$ (back-and-forth argument), and we may assume that $g(p) = 0$. $g$ induces a (bi-computable) homeomorphism $G: \mathbb{R} \to \mathbb{R}$.
Define $f = G\circ T\circ G^{-1}$.
Constructing the bijection:
We'll build a computable sequence of rationals $(q_n)_{n \in \omega}$ and define $g(T^n(p)) = q_n$.
Begin with $q_0 = 0$. Then compute enough of $p$ and $T(p)$ to determine how they are ordered ($p < T(p)$ or $T(p) < p$). If $p < T(p)$, choose $q_1$ to be a rational greater than 0; otherwise, choose $q_1$ to be a rational less than 0. In either case, choose $q_1$ to be the appropriate rational with the smallest Gödel number.
Then compute enough of $p$, $T(p)$ and $T^2(p)$ to determine how they are ordered, and choose $q_2$ to be a rational in the same relative position to $q_0$ and $q_1$. Again, choose the appropriate rational of smallest Gödel number.
Etc.
This is all a computable process, so $g$ is computable. It's a total order-preserving injection by construction.
Surjectivity is by induction on Gödel number. For a rational $r$, by the inductive hypothesis all rationals of smaller Gödel number are in the range of $g$. So fix an $n$ such that all rationals of smaller Gödel number occur in $q_0, \dots, q_{n-1}$, and assume $r$ does not occur in this list, as otherwise we are done. Define $C = \{i < n : q_i < r\}$. By the density of $X$, there is an $m \ge n$ with $T^i(p) < T^m(p)$ for all $i \in C$, and $T^m(p) < T^i(p)$ for all $i < n$ with $i \not \in C$. Fix the least such $m$. Then by construction, $q_m = r$.