What is the oxidation state of Boron in Ammonia Borane?
As per this source, the idea and definition of oxidation state is based on the following principle:


  The oxidation number of an atom in a molecule is based
  on a formalism that forces a covalent compound to possess
  complete ionic character and may be defined as the charge
  that an atom would have if all bonds to it are broken such
  that the ligands retain a closed-shell configuration ;
  an exception, however, refers to homonuclear bonds, in which
  case the bond is broken homolytically and a single electron
  is transferred to each atom.
  
  The oxidation number may
  thus simply be expressed as
  Oxidation number = charge on compound - charge on ligands
  


So,evidently, for the presented molecule $\ce{NH3BH3}$, we see that $\ce{NH3}$ is a ligand for the $\ce{BH3}$ moiety. So,you can heterolytically cleave the dative bond between $\ce{N}$ and $\ce{B}$ towards $\ce{N}$(as per the electronegativity trends),leaving no charge on the $\ce{B}$ as of now,as after this action,it's valence shell contains 3 electrons.Now,for finding oxidation state on $\ce{B}$,an interesting thing crops up due to the presence of the three $\ce{B-H}$ bonds. 
The article further enumerates:


  In many cases, the charges assigned to simple monoatomic
  ligands do not vary from compound to compound, as illus-
  trated by $\ce{F-}$, $\ce{Cl-}$, and $\ce{O^2-}$. However, a notable exception is
  provided by hydrogen for which both $\ce{H+}$ and $\ce{H-}$ have per-
  missible closed-shell configurations ($\ce{1s^0}$ and $\ce{1s^2}$
  , respectively).
  In this case, the charge assigned to hydrogen is determined by
  the relative electronegativity of the atom to which it is attached.


So again, due to sligthly higher electronegativity value of $\ce{H}$ than $\ce{B}$,the $\ce{H}$ atom becomes the ligand for the $\ce{B-H}$ bond. Hence,all the $\ce{B-H}$ heterolytically cleave towards $\ce{H}$,each cleavage leading to a +1 charge on $\ce{B}$ and -1 on $\ce{H}$. In totality,$\ce{B}$ ends up with +3 oxidation state, as it has lost all three electrons from it's valence shell.

Note: I highly recommend everybody to read that paper cited above. It's really insightful into the difference between valence,oxidation number and coordination number,which are often used interchangeably