I'm not entirely sure what you mean by 'as the base for a CRF representation'; continued fraction representations, at heart, don't use bases at all - just integers.
Assuming that what you mean is allowing the coefficients of a standard continued fraction representation to be elements of $\mathbb{Z}[\phi]$ rather than just $\mathbb{Z}$, then you run into trouble of a different source: in that case, there's no canonical CF representatation! The usual algorithm for generating a canonical continued fraction uses the floor operation $\lfloor x\rfloor$ (that is, 'the largest integer less than $x$') : $a_0 = \lfloor x\rfloor, a_1=\lfloor\frac{1}{x-a_0}\rfloor$, etc. But this operation can't be defined in $\mathbb{Z}[\phi]$ because the values in that ring are dense in the reals; there's no 'largest member less than $x$' for any $x$.
On the other hand, if you're interested in generalizing some of the properties of continued fractions to the ring of 'golden integers', there's another approach you could investigate:
As I'm sure you're already aware, the sequence of partial convergents to a continued fraction for some irrational number $x$ offers the best approximations to $x$, in a very canonical sense: each term $\frac{p}{q}$ in the sequence is the best approximation to $x$ of height less than $q$, where we define the height of a fraction $\frac{c}{d}$ in simplest terms to be just the denominator of the fraction. In essence, the height represents the simplicity of the number, and the best rational approximation property says that no 'simpler' number than any of the partial convergents of the CF can be a better approximation.
You could try and generalize this property by defining a canonical height for golden integers (for instance, $\mathrm{ht}(a+b\phi) = \max(|a|, |b|)$ or even just $\mathrm{ht}(a+b\phi) = |b|$) and then studying the sequence of 'convergents' to a number $x$ using this height and the best-approximation property; this is in some sense similar to the continued fraction approach in that both are looking at the best approximation to $x$ of the form $f(a,b)$ for particular functions $f$ and integers $a,b$ less than some specified bound (and studying how that approximation changes as the bound moves) — in the case of continued fractions then $f(a,b) = \frac{a}{b}$ while here it's $f(a,b) = a+b\phi$. It's very possible that the similarity of the problem means that these convergents have some canonical structure (similar to the way that the continued-fraction structure unites the best rational approximations) which you could use to define a continued-fraction equivalent for the representation by golden integers. Speaking personally, I'd certainly be curious to see whether anything interesting came out of it!
Best Answer
The golden ratio $$ x : 1 = 1 + x : x $$ leads to the equation $$ x^2 - x - 1 = 0 \quad (\#) $$ It can be transformed to two different equations of the form $$ x = F(x) $$ which then can be used to substitute the $x$ on the right hand side by $F(x)$ $$ x = F(F(x)) = F(F(F(x))) = \cdots $$
Your transformed version of $(\#)$ was this equation: $$ x = 1 + \frac{1}{x} \quad (*) $$
The repeated substitution of the RHS $x$ in $(*)$ with the RHS term $$ x \mapsto 1 + \frac{1}{x} $$ results in the expansion \begin{align} x &= 1 + \frac{1}{x} \quad (\mbox{EX}1.1) \\ &= 1 + \frac{1}{1 + \frac{1}{x}} \quad (\mbox{EX}1.2) \\ &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} \quad (\mbox{EX}1.3) \\ &= \cdots \end{align} and leads to the continued fraction $$ x = 1 + \frac{1\rvert}{\lvert 1} + \frac{1\rvert}{\lvert1} + \cdots \quad (\mbox{CF}1) $$
This continued fraction is positive (consisting only of additions and divisions of positive numbers) and will converge to the positive solution of $(\#)$.
Note that each of the steps $(\mbox{EX}1.n)$ is an equation equivalent to equation $(\#)$, having two solutions for $x$, while $(\mbox{CF}1)$ converges only to the positive root. This is because the continued fraction is the limit of these finite fractions: \begin{align} c_0 &= 1 \\ c_1 &= 1 + \frac{1}{1} = 2 \\ c_2 &= 1 + \frac{1}{1 + \frac{1}{1}} = 1.5 \\ c_3 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = 1.\overline{6} \\ c_4 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = 1.6 \\ \end{align}
Note that compared to the expansions $(\mbox{EX}1.n)$ the right hand side $1/x$ is dropped, which makes the difference.
To get the other root, the golden ratio equation $(\#)$ needs to be transformed into $$ x = \frac{1}{-1 + x} \quad (**) $$ Repeated substitution of the RHS $x$ in $(**)$ with the RHS term $$ x \mapsto \frac{1}{-1 + x} $$ results in the expansion \begin{align} x &= 0 + \frac{1}{-1 + x} \quad (\mbox{EX}2.1) \\ &= 0 + \frac{1}{-1 + \frac{1}{-1 + x}} \quad (\mbox{EX}2.2) \\ &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + x}}} \quad (\mbox{EX}2.3) \\ &= \cdots \end{align} leads to the continued fraction $$ x = 0 + \frac{1\rvert}{\lvert-1} + \frac{1\rvert}{\lvert-1} + \cdots \quad (\mbox{CF}2) $$ for the negative root of $(\#)$.
Note that all equations $(\mbox{EX}2.m)$ are equivalent to equation $(\#)$ and the equations $(\mbox{EX}1.n)$, thus have two solutions for $x$.
However the derived continued fraction $(\mbox{CF}2)$ is the limit of these finite fractions: \begin{align} d_0 &= 0 \\ d_1 &= 0 + \frac{1}{-1} = -1 \\ d_2 &= 0 + \frac{1}{-1 + \frac{1}{-1}} = -0.5 \\ d_3 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}} = -0.\overline{6} \\ d_4 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}}} = -0.6\\ \end{align} They result from dropping the right hand side $1/(-1+x)$ sub term.
Note that the negative root is nicely approached from above by the even numbered terms and from below by the odd numbered terms.