Cannizzaro reaction versus Nitroaldol reaction
Is the nitroaldol reaction that much faster?


My guess would be that the nitroaldol is faster than the Cannizzaro.  The Cannizzaro reaction involves


attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate
alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)
hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material


The nitroaldol involves


generation of the nitroalkyl carbanion
attack by the carbanion on the carbonyl to form a tetrahedral intermediate
capture of a proton by the negatively charged oxygen to produce a neutral molecule
elimination of water to produce a more stable (more conjugated) molecule than we started with (e.g. energetically downhill)


The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.


  Can the yields be so different due to the steric hindrance of the
  additional methyl group?


Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.


  What is the benefit of these nitrogen bases over $\ce{NaOH}$?


They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.