There is one thing that you can "see" when you have a surface embedded in $R^3$. Namelly, the moduli space of conformal (complex) structures on a surface of genus $g$, $g>0$ is non-compact. If you take in $R^3$ a surface that has a very long cylinder inside of it, such that the circle generated by it on the surface is not contractible, then in the moduli space of complex curves you get a point "close to its boundary". The longer the cylinder close you get to the boundary. En example will be a long and thin torus.
Considering such pictures you will be able to viualise at least points of the moduli space that are the most close to the boundary in the case of arbirtary genus.
Over thing you could destinguish are complex curves that admit an anti-holomorphic involution - just take a surface in $R^3$ symmetric with respect the $x,y$ plane.
As for small deformation of the sphere, unfotrunatelly I don't think it will be any easier to "see" that it is the same as on any other sphere. You will need PDE to prove it. Even to prove that it has a holomorphic sturcture you will need to know that for any metric on a 2-dimesnional surface you have local isotermal coordinates, where it looks as $f(x,y)(dx^2+dy^2)$f(x,y)(dx^2+dy^2)$.
ADDED.
By Pogorelov's theorem every sphere with positive Gauss curvature can be emdedded isometrically in R^3. So a sphere close to a unite sphere is just a sphere whose curvature is approximatively 1. Knowing this does not make it any easier for you to know that the conformal structure on this sphere is standard, according to what I understand.
