What happens when you apply the path integral to the Einstein-Hilbert action? The Einstein Field Equations emerge when applying the principle of least action to the Einstein-Hilbert action, and from what I understand the path integral formulation generalizes the principle of least action. What happens when you apply the path integral instead of the action principle to the Einstein-Hilbert action?
 A: Your questions essentially amounts to ask 

How do we quantize GR?

which is the starting point of quantum gravity (QG). GR is a non-renormalizable theory, at least from the traditional perspective of perturbation theory in QFT. So the path integral with the (exponentiated) Einstein-Hilbert action as weight factor cannot easily be used to make meaningful physical predictions. New approaches to QG are needed, such as e.g. string theory (ST).
A: The other answer and its comments are on point. I just wanted to add a little something. "What happens" is a kind of vague question. Like, what would you say "happens" when we put the action for a particle into the path integral?  In any case, I thought the question might partially be asking "what would it mean to do this?" In the case of QM, we integrate over the space of all possible paths for the particle; in QFT, we integrate over the space of all possible field configurations.  And quantum behavior arises from the fact that paths other than the classical one contribute something to the path integral - they happen in some sense, too. Thinking along these lines, what "happens" for the EH action in the path integral is that we now integrate over the space of all metrics - the "right" one (the one that solves the EFE) is no longer the only one that contributes to the path integral.
A: There are many issues associated with the path integral definition of the gravitational action, but here is one in particular : 
Path integrals tend to be rather ill defined in the Lorentzian regime for the most part, that is, of the form 
\begin{equation}
\int \mathcal{D}\phi(x) F[\phi(x)]e^{iS[\phi(x)]}
\end{equation}
Due to the fact that the integral oscillates, being a complex phase. To make them converge, a real factor is introduced, either by slight rotation of time ($t \rightarrow t(1 + i\varepsilon)$), or going all the way to Euclidian spacetime ($t \rightarrow it$). This gives the Euclidian path integral
\begin{equation}
\int \mathcal{D}\phi(x) F[\phi(x)]e^{-S_E[\phi(x)]}
\end{equation}
To converge properly, it is usually required that $S_E$ be positive. This is not the case for the gravitational action, which, in Euclidian form, is 
\begin{equation}
S_E = -\frac{1}{16\pi G} \int d^nx \sqrt{g} R_E(x)
\end{equation}
It can even be made arbitrarily negative since a conformal transformation involves a term of the form 
\begin{equation}
-\frac{6}{16\pi G} \int d^nx \sqrt{g} \Omega_{,a}(x) \Omega^{,a}(x)
\end{equation}
Since the conformal factor is more or less arbitrary, and you have to integrate over those configurations, it is a rather big problem to show convergence of the action. 
