Let $t = 1 + y^3$, we can rewrite $B\left(\alpha^2; \frac12, \frac13 \right)$ as
$$\int_0^{\alpha^2} t^{-1/2} (1-t)^{-2/3} dt
= \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{3y^2dy}{\sqrt{1+y^3} y^2}
= 3 \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{dy}{\sqrt{1+y^3}}$$
Following the setup in my answer to a related question. Let
$\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$ and $\wp(z)$ be the Weierstrass elliptic $\wp$ function with fundamental periods $1$ and $e^{i\pi/3}$. $\wp(z)$ is known to satisfy an ODE of the form
$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$
If one perform variable substitution $\;\displaystyle y = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$, one has
$$\frac{dy}{\sqrt{1+y^3}} = -dz\quad\text{ and }\quad
\begin{cases}
y\left(\frac{\sqrt{3}\eta}{3}\right)
= -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{3}\right) = 0\\
\\
y\left(\frac{\sqrt{3}\eta}{2}\right)
= -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2}\right) = -1
\end{cases}$$
Using this, we can express conjecture $(8)$ in terms of $y(\cdot)$ and/or $\wp(\cdot)$:
$$\begin{align}
& B\left(\alpha^2; \frac12, \frac13 \right)
\stackrel{?}{=} \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)} = \frac{\sqrt{3}}{4}\eta\\
\iff & 3\left[y^{-1}(-1) - y^{-1}(-\sqrt[3]{1-\alpha^2})\right]
\stackrel{?}{=} \frac{\sqrt{3}}{4}\eta\\
\iff & y^{-1}(-\sqrt[3]{1-\alpha^2})
\stackrel{?}{=} \frac{5\sqrt{3}}{12}\eta\\
\iff & \frac{4}{\eta^2}\wp\left(i\frac{5\sqrt{3}}{12}\right)
\stackrel{?}{=} \sqrt[3]{1-\alpha^2}
\end{align}
$$
Let $u_0 = i\frac{\sqrt{3}}{3}$, $u_{-1} = i\frac{\sqrt{3}}{2}$ and $u = i\frac{5\sqrt{3}}{12} = \frac12(u_0 + u_{-1})$. Using the addition formula for $\wp$ function,
we have
$$
\wp(2u) = \wp(u_0 + u_{-1})
= \frac14\left[\frac{\wp'(u_0)-\wp'(u_{-1})}{\wp(u_0)-\wp(u_{-1})}\right]^2 - \wp(u_0) - \wp(u_{-1})\\
=\frac14\left[\frac{-i\frac{\eta^3}{4} - 0}{0 - \frac{\eta^2}{4}}\right]^2 - 0 - \frac{\eta^2}{4}
= -\frac{\eta^2}{2}
$$
Using the duplication formula of $\wp$ function, we get
$$
-\frac{\eta^2}{2} = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u)
= \frac{9\wp(u)^4}{4\wp(u)^3-\frac{\eta^4}{16}} - 2\wp(u)
$$
Let $Y = \frac{4}{\eta^2}\wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to
$$\begin{align}
& Y^4 + 8 Y^3 + 8 Y - 8 = 0\tag{*1a}\\
\iff & Y(8+Y^3) = 8(1-Y^3)\tag{*1b}\\
\implies & (9-A^2)^3(1-A^2) = 512A^6\tag{*1c}\\
\iff & (A^4-24A^3+18A^2-27)(A^4+24A^3+18A^2-27) = 0\tag{*1d}
\end{align}$$
- Since $\alpha$ is a root for one of the factors in $(*1d)$, $A = \alpha$ satisfies $(*1d)$ and hence $(*1c)$.
- Since $0 < \alpha < 1$ implies $1 - \alpha^2 > 0$, $(*1c) \implies (*1b)$ in this particular case.
i.e. $Y = \sqrt[3]{1-\alpha^2}$ satisfies $(*1b)$ and hence $(*1a)$.
- Since $u$ lies between $u_0$ and $u_{-1}$, $\frac{4}{\eta^2}\wp(u) > 0$. Using
the fact $(*1a)$ has only one positive root, we find $\frac{4}{\eta^2}\wp(u) = \sqrt[3]{1-\alpha^2}$. i.e. conjecture $(8)$ is true.
Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector
$$\vec{y}(z)=\left(\begin{array}{c}
y_1 \\ y_2
\end{array}\right)=\left(\begin{array}{c}
_2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z)
\end{array}\right).\tag{1}$$
This is a single-valued vector function on $\mathbb{C}\backslash\{(-\infty,0]\cup[1,\infty)\}$. Its analytic continuation along a closed loop $\gamma$ gives rise to monodromy representation of $\pi_1(\mathbb{C}\backslash\{0,1\})$:
$$ \gamma\mapsto M_{[\gamma]},\qquad y(\gamma z)=M_{[\gamma]}y(z).$$
The monodromy group $G\subset GL(2,\mathbb{C})$ of the hypergeometric equation is generated by two matrices corresponding to simple loops around $0$ and $1$. In the case we are interested in these matrices are explicitly given by
$$M_0=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{-2\pi i /3}\end{array}\right),\qquad
M_1=C\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{2\pi i /3}\end{array}\right)C^{-1},\tag{2}$$
where the connection matrix $C=\left(\begin{array}{cc} 1 & 2^{\frac43} \\ -2^{\frac83} & 8\end{array}\right)$. If $G$ is finite, then $\vec{y}(z)$ has a finite number of branches, and moreover (Schwarz 1872), is algebraic.
It is not difficult to check that the monodromy group $G$ generated by $M_0$, $M_1$ from (2) is indeed finite. In particular, note that
$$M_0^3=M_1^3=I,\qquad M_1^2=-M_0M_1M_0, $$ $$M_1M_0M_1=-M_0^2,\qquad M_1M_0^2M_1=M_0^2M_1M_0^2.$$
It turns out that $G$ has order $24$ and is isomorphic to the binary tetrahedral group:
$$G\cong 2T=\langle s,t\,|\,(st)^2=s^3=t^3\rangle, $$
where the generators can be identified as $s=M_0M_1M_0M_1M_0$, $t=M_1M_0M_1M_0M_1$.
Corollary: The hypergeometric functions in (1) are algebraic.
Algebraic solutions of the hypergeometric equations are classified by the so-called Schwarz table, and have been studied by many mathematicians, see e.g. the bibliography in this paper by R.Vidunas. Their explicit construction is somewhat involved but relatively straightforward - at least when the corresponding algebraic curve has genus $0$ (the genus can be determined independently from the Riemann-Hurwitz formula).
In our case the task simplifies even more as our parameter values can be obtained from the genus $0$ tetrahedral formula (2.4) of the above mentioned paper by a combination of a linear trasformation (sending $\frac56$ to $\frac43-\frac56=\frac12$) and differentiation (transforming $\frac43$ into $\frac13$). The result is
$$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}.$$
Corollary: The antiderivative $\displaystyle\int \mathcal{R}\left(x,y(x)\right)dx$, where $y(x)={}_2F_1\left(\frac16,\frac12;\frac13;-x\right)$ and $\mathcal{R}(x,y)$ is rational in both arguments, can be expressed in terms of elementary functions.
Example:
The transformation $r\mapsto x(r)=\frac{r(r+2)^3}{(r+1)(1-r)^3}$ bijectively maps $(0,1)$ to $(0,\infty)$, and therefore the initial integral becomes
\begin{align}\mathcal{I}&=\int_0^1 \left(\frac{\sqrt{1-r^2}}{2r+1}\right)^{12}\left(\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)'dr=\\&=
2\int_0^1 \frac{(1+r)^4(1-r)^2(r+2)^2}{(2r+1)^{10}} dr=\\&=\frac{80\,663}{153\,090}.
\end{align}
Best Answer
The conjecture is true, as are the other cases reported in the comments where $f(z) := {}_2F_1 \left( \frac13, \frac13; \frac56; z \right)$ takes algebraic values for special rational values of $z$. There are a few others obtained from the symmetry $z \leftrightarrow 1-z$ (these ${}_2F_1$ parameters correspond to a hyperbolic triangle group with index $6,6,\infty$ at $c=0,1,\infty$, so the $z=0$ and $z=1$ indices coincide); e.g. $f(-1/3) = 2 / 3^{2/3}$ pairs with $f(4/3) = 3^{-2/3} (5-\sqrt{-3})/2$. ($z=1/2$ pairs with itself, and the pair $f(-4)$ and $f(5)$ has been noted already; the OP's $f(-27) = -4/7$ pairs with $f(28) = \frac12 - \frac3{14} \sqrt{-3}$.) Somewhat more exotic are $$ f\big({-}4\sqrt{13}\,(4+\sqrt{13})^3\big) = \frac7{13\,U_{13}}\\ f\big({-}\sqrt{11}\,(U_{33})^{3/2}\big) = \frac{6}{11^{11/12}\, U_{33}^{1/4}}, $$ with fundamental units $U_{13}=\frac{3+\sqrt{13}}2,\;U_{33}=23+4\sqrt{33}$ and further values at algebraic conjugates and images under $z \leftrightarrow 1-z$.
In general, for $z<1$ the integral formula for $f(z)$ relates it with $$ \int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} } $$ which is half of a "complete real period" for the holomorphic differential $dx/y$ on the curve $C_z : y^6 = (1-x)^3 x^4 (1-zx)^2$. This curve has genus $2$, but is in the special family of genus-$2$ curves with an automorphism of order $3$ (multiply $y$ by a cube root of unity), for which both real periods are multiples of the real period of a single elliptic curve $E_z$ (a.k.a. a complete elliptic integral). In general the resulting formula doesn't simplify further, but when $E_z$ has CM (complex multiplication) its periods can be expressed in terms of gamma functions. For $z = -27$ and the other special values listed above, not only does $E_z$ have CM but the CM ring is contained in ${\bf Z}[\rho]$ where $\rho = e^{2\pi i/3} = (-1+\sqrt{-3})/2$. Then the $\Gamma$ and $\pi$ factors of the period of $E_z$ exactly match those in the integral formula, leaving us with an algebraic value of $f(z)$. It turns out that the choice $z = -27$ makes $E_z$ a curve with complex multiplication by ${\bf Z}[7\rho]$. The others from the comments lead to ${\bf Z}[m\rho]$ with $m=1,2,3,5$, and the examples where $z$ is a quadratic irrationality come from ${\bf Z}[13\rho]$ and ${\bf Z}[11\rho]$.
One way to get from $C_z$ to $E_z$ is to start from the change of variable $u^3 = (1+cx)/x$, which gives $$ f(z) = \int_{\root 3 \of {1-z}}^\infty \frac{3u \, du}{\sqrt{(u^3+z)(u^3+z-1)}}. $$ and identifies $C_z$ with the hyperelliptic curve $v^2 = (u^3+z)(u^3+z-1)$. Now in general a curve $v^2 = u^6+Au^3+B^6$ has an involution $\iota$ taking $u$ to $B^2/u$, and the quotient by $\iota$ is an elliptic curve; we compute that this curve has $j$-invariant $$ j = 6912 \frac{(5+2r)^3}{(2-r)^3(2+r)} $$ where $A = rB^3$. (There are two choices of $\iota$, related by $v \leftrightarrow -v$, and thus two choices of $j$, related by $r \leftrightarrow -r$; but the corresponding elliptic curves are $3$-isogenous, so their periods are proportional.) In our case $r = A/B^3 = -(2z+1)/\sqrt{z^2+z}$ (in which the $z \leftrightarrow 1-z$ symmetry takes $r$ to $-r$). Taking $z=-27$ yields $j = -2^{15} 3^4 5^3 (52518123 \pm 11460394\sqrt{21})$, which are the $j$-invariants of the ${\bf Z}[7\rho]$ curves; working backwards from the $j$-invariants of the other ${\bf Z}[m\rho]$ curves we find the additional values of $z$ noted in the comments and earlier in this answer.