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.
Best Answer
If you allow the functions to be constant on some intervals, then there are some easy examples, and Ricky has provided one.
But if you rule that out, then there can be no examples, even with countably many functions. To see this, suppose that $f_n$ is a list of countably many continuous functions which are never constant on an interval. Enumerate the pairs $(r,n)$ of rational numbers $r$ and natural numbers $n$ in a countable list $\langle (r_0,n_0), (r_1,n_1),\ldots\rangle$. Let $C_0$ be any closed interval. If the closed interval $C_i$ is defined, consider the function $f_{n_i}$ and the rational value $r_i$. Since $f_{n_i}$ is not constant value $r_i$ on $C_i$, we may shrink the interval to $C_{i+1}\subset C_i$ such that $f_{n_i}$ on $C_{i+1}$ is bounded away from $r_i$. By compactness, there is some $x\in C_i$ for all $i$. Thus, $f_n(x)$ is not $r$ for any rational number $r$.