how to determine the number of characters in the argument of a command

It's easy with expl3:

\documentclass{article}
\usepackage{xparse} % loads expl3

\ExplSyntaxOn

\NewDocumentCommand{\mathvar}{m}
 {
  \int_compare:nTF { \tl_count:n { #1 } > 1 }
   {
    \mathrm{#1}
   }
   {
    #1 % or \mathit{#1}, but I wouldn't do it
   }
 }

\ExplSyntaxOff

\begin{document}

$\mathvar{V}\ne\mathvar{Var}$

\end{document}

The code should be self-explaining: \tl_count:n counts the number of items in its argument (ignoring spaces, however).

I wouldn't use \mathit for single letter variables, as normal math italic is specially tailored for the purpose.

An implementation with xstring:

\documentclass{article}
\usepackage{xstring}

\newcommand{\mathvar}[1]{%
  \begingroup\noexpandarg
  \StrLen{#1}[\temp]%
  \ifnum\temp>1
    \mathrm{#1}%
  \else
    #1%
  \fi
  \endgroup
}

\begin{document}

$\mathvar{V}\ne\mathvar{Var}$

\end{document}

Here's a LuaLaTeX-based solution. It defines a macro called \numchars, which returns the number of characters in its argument, and an implementation of \mathvar. Care is taken not to count whitespace characters in the argument of \mathvar.

% !TEX TS-program = lualatex
\documentclass{article}    
\usepackage{luacode} %  for "\luastring" macro and "luacode" environment
\begin{luacode}
function numchars ( s )  -- disregard any whitespace characters
   return tex.sprint ( unicode.utf8.len ( unicode.utf8.gsub ( s , "%s", "")))
end 
\end{luacode}
\newcommand\numchars[1]{\directlua{numchars(\luastring{#1})}}
\newcommand\mathvar[1]{%
   \ifnum\numchars{#1}<2\mathit{#1}\else\mathrm{#1}\fi}

\begin{document}
$\mathvar{ V }$, $\mathvar{VaR}$
\end{document}

Plainer solution.

\documentclass{scrartcl}

\long\def\gobbleone#1{}
\newcommand*\var[1]{\if\relax\detokenize\expandafter{\gobbleone#1}\relax#1\else\mathrm{#1}\fi}

\begin{document}

$\var{A} = \var{B} = \var{Whatever}$

\end{document}