Dear Igor,
Just to add some context to the answers that you have received: what you are considering is (a local version of) the following geometric situation: a map $X \to Y,$ where $Y$ is smooth, $X$ is normal, and the fibres are finite. Now by Noether normalization (which is just an algebraic way of describing projection to a hyperplane) any variety can be made finite over a smooth one, and in particular, this is true of any normal variety.
Thus your question is (essentially) ``is every normal variety smooth?'', to which the answer is no as soon as the dimension is $> 1$. This is where Angelo's answer comes from: he wrote down the simplest isolated singularity on a surface in 3-space (the vertex of a cone), and this singularity is normal but singular. (Normal implies that the singularities are in codimension 2, and for a hypersurface in an affine or projective space, the converse holds.)
To have a positive answer in some situation, you need to control the branching of your finite map in some way (it along the branch locus that singularities can appear), and the references that BCnrd mentions deal with this kind of question.
One more remark, which you probably know (but just to be sure): if $A$ is a DVR (i.e. if you are in the one dimensional case), then $B$ will be regular (assuming some reasonable background hypotheses, so that $B$ is finite over $A$, for example); geometrically, for a curve, being normal is the same as being smooth.