Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$

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.

(This is more a comment than answer, but I couldn't get MathJax to properly show it in comments)

Here is a nice identity (equation (21) of this paper with $x=-1/7$): $$_2F_1 \left(a,a+\frac{1}{2};\frac{4a+5}{6};-\frac{1}{7}\right)=\left(\frac{7}{4}\right)^a {_2}F_1 \left(\frac{a}{3},\frac{a+1}{3};\frac{4a+5}{6};-27\right)$$

It's an example of a cubic transformation. Possibly, one can at this point use contiguous relations to make some progress.

Regarding your secondary question, by appealing to the classical j-function at defined arguments, it seems there are infinitely many algebraic numbers $z$ such that the $_2F_1$ evaluates to an algebraic number. Some examples, $$_2F_1\left(\frac13,\frac13;\frac56;-z_1\right)= \frac9{17} \big(833+324\cdot17^{1/3}-252\cdot17^{2/3}\big)^{1/6}$$ $$2F_1\left(\frac13,\frac13;\frac56;-z_2\right)= \frac{10}{3\cdot19} \big(2+2\cdot19^{1/3}-19^{2/3}\big)$$ where, $$z_1 =4\big(19894+7737\cdot17^{1/3}+3009\cdot17^{2/3}\big)$$ $$z_2 =\frac{1}{3}\big(1464289+548752\cdot19^{1/3}+205648\cdot19^{2/3}\big)$$ See also this post.