Given, the unit interval $I$ endowed with the Lebesgue measure $\mu$, and let $A$ be the (Boolean) algebra of Jordan measurable subsets of $X$ with respect to $\mu$, (i.e. those sets that satisfying $\mu(\partial(A))=0)$, modulo sets of zero measure.
Is it true that every (nontrivial) (Boolean) subalgebra $B$ of $A$ is isomorphic to $A$.
The answer to your question, as we have already stated in comments, is negative. A nonatomic counterexample is the algebra $B$ generated by intervals of the form $[0,1/n]$ where $n$ is a nonnegative integer. This algebra is countable, while $A$ is uncountable -- every two different intervals $(a,b)$ and $(c,d)$ in $I$ have the symmetric difference of nonzero Jordan measure, hence they are in different equivalent classes after division by the ideal of Jordan null sets. Thus $B$ and $A$ are nonisomorphic.
So far I have encountered only one study of the algebra $A$, or more precisely the algebra $\mathscr{J}$ of all Jordan-measurable sets in $I$ (ie. without dividing by null sets). It was a paper by Schachermayer entitled On some classical measure-theoretical theorems for non-complete Boolean algebras, published in the Dissertationes Mathematicae series in 1982. The paper is rather difficult to read, as it requires some knowledge about vector measures and $C(K)$-spaces. The result in the article states that $\mathscr{J}$ has the Nikodym Boundedness Property and Orlicz-Pettis Property but it fails to have the Grothendieck Property.
If you look for some informations on Jordan measure just read these excelent notes by Terrence Tao: http://terrytao.wordpress.com/2010/09/04/245a-prologue-the-problem-of-measure/.