The recent answer to an old question of mine made me aware of The Ramanujan Machine. So it seems like so far, the number of continued fraction representations for $\zeta(3)$ of this polynomial kind is indeed very limited/sporadic (but of course one never knows for sure, as there are just soo many possibilities, see below).
Now the same site has a number of similar (still conjectural) continued fraction representations for the Catalan constant. And in fact, I have noticed that these are not at all sporadic but belong to two infinite families. The "numerator polynomials" $b_n$ are all of the form $b_n=-2n(n+i)(n+j)(n+k)$ with $i,j,k\in\mathbb N_0$.
Those of them for which the "denominator polynomials" $a_n$ are unique in the table all have the property that $i\equiv j\equiv k\pmod 2$, and if we take for such a triple $(i,j,k)$ $$a_n:=3n^ 2 + [2(i+j+k)+3]n + [(i+1)(j+1)(k+1)-ijk],$$ there is strong numerical evidence that the corresponding continued fraction has the closed form $$f_{i,j,k}:=a_0+\frac{b_1}{a_1+\frac{b_2}{\ddots}}=\frac1{pG+q },$$ where (for even $i,j,k$ as well as for odd $i,j,k$) $p$ and $q$ are rationals, and more precisely $$p=\frac{(-1)^i\sqrt{\pi}^{3}~ \Gamma(i+1) \Gamma(j+1) \Gamma(k+1)}{ 2^{i+j+k+1} \Gamma(\frac{i + j + 1}2) \Gamma(\frac{i + k + 1}2) \Gamma(\frac{j+k + 1}2)}=\frac{2^{i+j+k-1} ( \frac{i + j}2)!~( \frac{i + k}2)!~( \frac{j+k}2)!~i!~j!~k!}{(-1)^i(i+j)!~(i+k)!~(j+k)!}.$$
It is not hard to read off (but don't forget, it is still all only conjectural!) the general recursion formula, for even and odd alike, $$\frac{k(k-1)}{f_{i,j,k}}=\frac{(k+i-1)(k+j-1)}{f_{i,j,k-2}}-1.$$ Remembering that $f_{i,j,k}$ is symmetric in $i,j,k$, this allows to reduce $f_{i,j,k}$ to $\dfrac{1}{f_{0,0,0}}=2G$ for even $i,j,k$ and to $\dfrac{1}{f_{1,1,1}}=-2G+2$ for odd $i,j,k$. So it is possible to write the general form of $q$ as well, but it will be messy, e.g. for $i=j=1$ and $k=1,3,5,7,9,11$ we have already $q=2,\frac {17}{3!},\frac {419}{5!},\frac {20411}{7!},\frac {1648251}{9!},\frac {199075491}{11!}$. (But see below!)
The remaining formulas in the table, i.e. where several $b_n$'s share the same $a_n$, belong to a different family, more precisely a family of families. In all those cases, at least two of $i,j,k$, say $i$ and $j$, are even. Fixing a pair of even $i,j\in2\mathbb N_0$, the polynomial $$a_n:=3n^ 2 + (2i+2j+3)n + (i+1)(j+1)$$ works for all $k\in\mathbb N_0$, including the odd ones, and yields a continued fraction with the closed form (for convenience of notation, writing here $2i$ and $2j$ instead) $$g_{2i,2j,k}:=a_0+\frac{b_1}{a_1+\frac{b_2}{\ddots}}=\frac1{pG+q },$$ where for all $i,j\in\mathbb N_0$, $$p=\frac{\sqrt{\pi}^{3}~\Gamma(i+1)\Gamma(j+1)\Gamma(k+1) \Gamma(k-i-j + \frac{1}2)}{2 \Gamma(i + \frac{1}2) \Gamma(j +\frac{ 1}2) \Gamma(k-i +\frac{ 1}2) \Gamma(k-j +\frac{ 1}2)}. $$
Putting it all in terms of factorials is again possible but not practical here, because if $k<i+j$, we would have to replace undefined fractions of type $\frac{( -m)!}{(-2m )!}$ by $(-1)^{m}\frac{ m!}{(2m )!}$.
So, switching now back to the original notation, $g_{i,j,k}$ is only defined for even $i,j$ and symmetric in $i\leftrightarrow j$.
For these families of the second kind, there are similar recursion formulas for the continued fraction $g_{i,j,k}$. In fact it appears that for fixed even $i,j$ $$\frac{2k(2k-i-j-1)}{g_{i,j,k}}=\frac{(2k-i-1)(2k-j-1)}{g_{i,j,k-1}}-1.$$ For $g_{i,j,0}$ there is a recursion of similar type to reduce $i$ and $j$, $$\frac j{g_{i,j,0}}=\frac {i+j-1}{g_{i,j-2,0}}-\frac1{j-1}.$$ Knowing $g_{0,0,0}=\dfrac1{2G}$, someone with a lot of courage may work out the general case here too, very messy but still doable. I have done it just for $i=j=0$, where we get $$\frac{1}{g_{0,0,k} }= \frac 1{2^{2k-1}}\binom{2k}k\left(G-\sum_{t=1}^k \frac{4^{t-1}}{ t(2t-1) \binom{2t}t}\right). $$
Hopefully, these precise formulas may allow someone more knowledgeable about continued fractions to progress towards a proof. Maybe it is possible to prove the recursion formulas just by playing around with the continued fractions, which would be a first step and then "only" leave to prove the continued fractions for $f_{0,0,0} (=g_{0,0,0})$ and $f_{1,1,1}$.
Note that if $i,j,k$ are all even, the numerators $b_n=-2n(n+i)(n+j)(n+k)$ can be combined with up to four different polynomials $a_n$: One of the first kind above, and for the second kind up to three (if $i,j,k$ are all distinct), one for each fixed pair in $\{i,j,k\}$. Otherwise stated, while we have $g_{i,j,k} =g_{j,i,k}$, we will get different polynomials $a_n$ for other permutations.
To come back to the initial paragraph: Given that $G$ and $\zeta(3)$ are in a certain sense not too different in nature, I'd not be surprised if there are infinite families of continued fraction representations of similar kinds not for $\zeta(3)$ alone (like there are probably not many for $G$ alone), but for $\dfrac1{p\zeta(3)+q}$ where $p,q$ are rational. Let's hit the beehive!
