I) Recall that the path integral formulation comes in (at least) two versions: Lagrangian & Hamiltonian. It is often argued that the Hamiltonian version is more fundamental, cf. e.g. this Phys.SE post.
II) On one hand, the Dirac-Bergmann analysis/singular Legendre transformation
yields that the Nambu-Goto (NG) Hamiltonian Lagrangian density reads
$${\cal L}_{NG,H} ~:=~ P\cdot \dot{X}-{\cal H}, \qquad {\cal H}~=~\lambda^{\alpha} \chi_{\alpha}, \qquad \alpha~\in~\{0,1\},\tag{1}$$
with two first class constraints
$$ \chi_0~:=~P\cdot X^{\prime}~\approx~0, \qquad \chi_1~:=~\frac{P^2}{2T_0}+\frac{T_0}{2}(X^{\prime})^2~\approx~0,\tag{2} $$
see e.g. this Phys.SE post.
Here $\lambda^0$ and $\lambda^1$ are Lagrange multipliers for the first class constraints (2).
III) The original Nambu-Goto square root Lagrangian density
$$\begin{align} {\cal L}_{NG}~:=~&-T_0\sqrt{{\cal L}_{(1)}}, \cr {\cal L}_{(1)}~:=~&-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta}
~=~(\dot{X}\cdot X^{\prime})^2-\dot{X}^2(X^{\prime})^2~\geq~ 0, \end{align}\tag{3}$$
is re-obtained by integrating out the variables $P$, $\lambda^0$ and $\lambda^1$ in the Nambu-Goto Hamiltonian Lagrangian density (1) if we restrict the variable
$$\lambda^1~>~0 \tag{4} $$
to be positive. The ineq. (4) is needed in order to exclude an unphysical negative square root branch. Note that the ineq. (4) implies that the corresponding first-class constraint $\chi_1$ is technically speaking not enforced with a Dirac delta distribution in the path integral.
IV) On the other hand, the Polyakov (P) De Donder-Weyl (DDW) Lagrangian density reads
$$\begin{align}{\cal L}_{P,DDW}~=~&P^{\alpha} \cdot \partial_{\alpha}X
+\frac{\gamma_{\alpha\beta}P^{\alpha}\cdot P^{\beta}}{2T_0\sqrt{-\gamma}} \cr
~=~& P^{\tau}\cdot \dot{X} +P^{\sigma}\cdot X^{\prime}+ \frac{(P^{\sigma} + \lambda^0 P^{\tau})^2}{2T_0 \lambda^1} - \frac{\lambda^1}{2T_0} (P^{\tau})^2, \end{align}\tag{5}$$
where we have introduced a world-sheet (WS) metric $\gamma_{\alpha\beta}$ and polymomenta $P^{\alpha}=(P^{\tau};P^{\sigma})$. See also e.g. my Phys.SE answer here.
V) In the second equality of eq. (5), we have defined two auxiliary variables, $\lambda^0$ and $\lambda^1$, as follows:
$$\begin{align} \lambda^0~=~&\frac{\gamma_{\tau\sigma}}{\gamma_{\sigma\sigma}}
~=~-\frac{\gamma^{\tau\sigma}}{\gamma^{\tau\tau}}, \cr
\lambda^1~=~&\frac{\sqrt{-\gamma}}{\gamma_{\sigma\sigma}}
~=~ \frac{-1}{\sqrt{-\gamma}\gamma^{\tau\tau}}~\geq~0
\quad\Leftrightarrow\quad\gamma~:=~\det\left(\gamma_{\alpha\beta}\right)_{\alpha\beta}~=~-\left(\lambda^1\gamma_{\sigma\sigma}\right)^2~\leq~0 . \end{align} \tag{6}$$
The sign of the variable $\lambda^1$ must be positive in order to make the kinetic term in the Polyakov Lagrangian density positive definite. See also my Phys.SE answer here for more details.
Note the three WS metric components $\gamma_{\tau\tau}$, $\gamma_{\tau\sigma}$, and $\gamma_{\sigma\sigma}$ only enter the Polyakov De Donder-Weyl Lagrangian density (5) via the two combinations $\lambda^0$ and $\lambda^1$! (This neat fact is connected to Weyl symmetry.)
VI) To achieve the Polyakov Hamiltonian Lagrangian density ${\cal L}_{P,H}$ from the Polyakov De Donder-Weyl Lagrangian density (5), we should keep the momentum variable $P^{\tau}\equiv P$, and integrate out the auxiliary variable $P^{\sigma}$.
It turns out that the Polyakov Hamiltonian Lagrangian density becomes precisely the Nambu-Goto Hamiltonian Lagrangian density (1)!
Note that this off-shell conclusion is stronger than the usual claim that the Polyakov string and the Nambu-Goto string have the same classical EOMs on-shell!
Altogether, the only remaining difference is in the third degree of freedom of the WS metric, which corresponds to the Weyl symmetry, and whose path integral contribution naively factorizes and hence decouples. A more careful analysis reveals regularization issues for both the Polyakov string and the Nambu-Goto string, which potentially leads to a conformal anomaly. In a flat target space (TS), the conformal anomaly famously only vanishes in the critical dimension $D=26$.
TL;DR: In conclusion, since the Hamiltonian Lagrangian density of the Nambu-Goto string and the Polyakov string are identical, then any path integral quantization scheme (that is consistent with the Hamiltonian formulation) must be identical as well.
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