The first part of this answer is for an earlier version of the question, with $x_0$ a rational represented by an arbitrarily large bitstring.
There exists functions $f$ such that no algorithms can predict the next bit of the output sequence.
A simple example is $f(x)=\begin{cases}2x&\text{if }x<1/2\\2x-1&\text{otherwise}\end{cases}$.
With this function, the binary sequence produced is the¹ binary representation of $x_0$ (starting at the first bit after decimal point), which can't be predicted. Is that $f$ "a chaotic map"? I can't tell.
We can make $f$ continuous, e.g. $f(x)=\begin{cases}2x&\text{if }x<1/2\\2-2x&\text{otherwise}\end{cases}$.
The relation between a binary representation of $x_0$ and the sequence remains such that changing the $i^\text{th}$ bit of that binary representation of $x_0$ changes the $i^\text{th}$ bit of the sequence. I think I have seen this function, or a close cousin, dubbed "a chaotic map".
We can make $f$ indefinitely derivable with $f(x)=\frac{43}{11}\,x\,(1-x)$ (a case of "the logistic map with rational parameter", and "a chaotic map" by most accounts). Without proof: for any bitstring in $\{0,1\}^{n+1}$ there exists an $x_0$ such that the first $n+1$ bits output are that bitstring, thus the next bit is not predictable with certainty.
Now for the revised question with
for any $n$, even when $n >|x_0|$ where $|x_0|$ is the length of the bit-string representing $x_0$.
Without proof: with $f(x)=\frac{43}{11}\,x\,(1-x)$ and most $x_0$, the question's generator requires work exponential in the number of bits produced (argument: $f^n(x)$ is a polynomial of degree $2^n$, thus it seems that evaluating it to a give accuracy requires knowing $x$ with an exponential number of bits). Thus the question's generator does not meet the standard criteria of being a Polynomial-Time algorithm, and thus does not meet the standard definition of a PRG, regardless of predictability. At least, the cost and memory requirement grows so fast that it's not useful in practice.
On the other hand, for most fixed $x_0$ (perhaps, all those that do not make the bit sequence generated ultimately periodic), it's possible to make a partial predictor. In particular, the output sequence is sizably biased towards $1$. Thus a distinguisher is much easier than computing the sequence. I think that simple fixes like changing the threshold $1/2$ to the expected mean still will allow a polynomial-time distinguisher simply by computing the frequency of sequences of a number of bits.
¹ For numbers with two binary representations, that is of the form $a/2^k$, take the lexicographically first: e.g. $3/4$ is $.1011111111111111111…$
