Define $\color{blue}{q=e^{2\pi i \tau}}$ and Dedekind eta function $\eta(\tau)$. Note: I found these relations empirically, but their consistent forms suggest they can be rigorously proven.
I. $p=2$
We have,
$$\left(\frac{8}{\alpha^8+8}\right)^2+\left(\frac{\beta^8}{\beta^8+32}\right)^2=1\tag1$$
$$\alpha=\frac{\eta(\tau/2)}{\eta(2\tau)},\quad\beta=\frac{\eta(\tau)}{\eta(4\tau)}$$
If $\tau=\sqrt{-n}$, then,
$$\frac{_2F_1\Big(\tfrac14,\tfrac34;1;\,1-\big(\tfrac{8}{\alpha^2+8}\big)^2\Big)}{_2F_1\Big(\tfrac14,\tfrac34;1;\,\big(\tfrac{8}{\alpha^8+8}\big)^2\Big)}=\color{red}{\sqrt{2n}}$$
II. $p=3$
We have,
$$\left(\frac{3}{\gamma^3+3}\right)^3+\left(\frac{\delta^3}{\delta^3+9}\right)^3=1\tag2$$
$$\gamma=\frac{\eta(\tau/3)}{\eta(3\tau)},\quad\delta=\frac{\eta(\tau)}{\eta(9\tau)}$$
If $\tau=\sqrt{-n}$, then,
$$\frac{_2F_1\Big(\tfrac13,\tfrac23;1;\,1-\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\,\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}=\color{red}{\sqrt{3n}}$$
Note: The identity,
$$\left(\frac{x^3y + y}{x y^3 + x}\right)^3+\left(\frac{x^3 – y^3}{x y^3 + x}\right)^3=1,\quad \text{if}\; x^3+y^3=1$$
should guarantee infinitely many eta parametrizations to $(2)$. Note also that,
$$\gamma^3+3=4C^2(q)+\frac1{C(q)}\\
\delta^3+3=4C^2(q^3)+\frac1{C(q^3)}$$
with cubic continued fraction,
$$C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}=\cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}$$
$\color{blue}{Update:}$
I just realized that the cubic eta identity $(2)$ is equivalent to the Borweins' cubic theta identity,
$$\frac{c^3(q)}{a^3(q)}+\frac{b^3(q)}{a^3(q)} = 1$$
where,
$$a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$
$$b(q) = \tfrac{1}{2}\big(3a(q^3)-a(q)\big)$$
$$c(q) = \tfrac{1}{2}\big(a(q^{1/3})-a(q)\big)$$
from Ramanujan's Notebooks Vol. V, p.93.
III. $p=4,8$
$$\left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4+\left(\frac{\vartheta_4(0,q)}{\vartheta_3(0,q)}\right)^4 = 1\tag{3a}$$
with Jacobi theta function $\vartheta_n(0,q)$. If $q=e^{2\pi i \tau}$, then equivalently,
$$\left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8+\left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1\tag{3b}$$
hence the addends of $(3a)$ and $(3b)$ are equal. Expressed as, $$U^8(\tau)+V^8(\tau) =1$$
If $\tau=\sqrt{-n}$, then,
$$\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-U^8(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,U^8(\tau)\big)}=\color{red}{\sqrt{4n}}$$
This has a beautiful octic continued fraction studied by Ramanujan,
$$U(\tau)=\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}= \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$
and can solve the general quintic.
IV. Question
Q: Are there analogous eta quotients to parameterize the Fermat quintic $x^5+y^5=1$? And can we also use the Rogers-Ramanujan cfrac to do so?
Best Answer
(Courtesy of a comment by Nemo who suggested Huber's paper.)
In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ and the Rogers-Ramanujan continued fraction, $$R(z)=q^{1/5}\frac{H(q)}{G(q)}=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$
then Huber's four functions in simplified form are,
$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ as well as $c,d$ are Rogers-Ramanujan cfracs, $$\frac{a(\tau)}{b(\tau)} = R(\tau)\tag3$$ $$\frac{c(\tau)}{d(\tau)} = \phi\, R\big(\tfrac{-1}{5\tau}\big)\tag4$$ so $R(\tau)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(5\tau)}{\phi+R(5\tau)}\tag5$$
It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. This is given by,
$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag6$$
where, $$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ and which can be derived by combining $(1),(2)$.