Can Schwinger effect occur in electric field of a nucleus? I know that electron-positron pairs can be produced within a uniform electric field with electric field strength above $10^{18}V$. Electric field of such strength can be found near atomic nuclei. So if one has an atomic nucleus free of its electrons, can electron-positron pairs be produced at it's surface?
 A: If you have a naked (no electrons near it) charged nucleus with $Z= 137 $ then the Dirac equation predicts that  the binding energy for a single electron becomes equal to $-m_{\rm e}c^2$. Beyond that value of $Z$  the Dirac equation has no real-energy solution. The usual accounts take this to mean that  an electron-positron pair will be produced, the positron escapes to infinity and the electron falls into the nucleus where it  converts a proton to a neutron so    $Z$ is reduced by one. The process continues until $Z<137$ and the vacuum becomes stable again. See here
So the answer to your question is "yes" in theory, but no nuclei with $Z>137$ are available for experimental study.
Added material: I am amending my answer  slighltly after further reading.  That no solutions exist  to the Dirac equation with a point charge $Z>137$ is due to the singular character of the Coulomb potential of the  point-charge Dirac equation. This is a mathematical pathology  of the point-charge idealization, and not a physical instability. At that point  the lowest energy state has $E=0$ ( i.e $-m_{\rm e}c^2$ below the lowest energy of a free electron). An electron can appear and   occupy the zero-energy state with no energy cost -- but that electron has to come from somewhere and the resulting positron costs energy $+m_{\rm e}c^2$ --- so that the vacuum  is not yet unstable to pair creation.
If one has an extended charge distribtion due to a finite size nucleus the "falling into the center" pathology of   the Dirac equation is obviated, and the bound state energies can get below $E=0$ and reach $E=-m_{\rm e}c^2$. This is the top of the negative-energy spectrum for the free electron and so now one can produce electron-positron pairs at no cost. The electron will bind in the lowest energy state of the Dirac equation and the positron will escape to infinity.
