Is it true that every real closed field can be elementarily embedded in some other real closed filed with the same Archimedean classes (I mean in a proper extension)? Can for example real numbers be elementarily embedded in another real closed field with the same Archimedean classes? R (real numbers) is not the only Archimedean field, is it right?

The question is somewhat ambiguous, it’s not clear whether the Archimedean classes are meant to be additive or multiplicative. I will assume the former, i.e., equivalence classes of the relation $$a\sim b\Leftrightarrow\mathrm{sign}(a)=\mathrm{sign}(b)\land\exists n\in\omega\smallsetminus\{0\}\\,(n^{1}a\leb\le na).$$ First, since realclosed fields (rcf) have elimination of quantifiers, any embedding between them is automatically elementary. Thus the question is whether every rcf $R$ has a proper rcf extension $S$ with the same Archimedean classes (i.e., every $s\in S$ is $\sim$ to some $r\in R$). As Andreas noted above, this property does not hold in general, and in particular, $\mathbb R$ has no proper Archimedean extension. On the other hand, it holds for many other realclosed fields: for example, any Archimedean rcf different from $\mathbb R$ has a proper Archimedean rcf extension (namely, $\mathbb R$). I think the following characterization holds: Proposition: If $R$ is a rcf, the following are equivalent:
On the one hand, let $S\supseteq R$ be a rcf with the same Archimedean classes and $x\in S\smallsetminus R$. We can assume $x>1$. There exists $c\in R$ such that $c\sim x$; WLOG $c< x< 2c$. Then $0< x/c1< 1$, and the Dedekind cut on $R$ determined by $x/c1$ is easily seen to satisfy $(*)$. On the other hand, assume the cut $\langle A,B\rangle$ is given. We define an ordering on the rational function field $F=R(x)$ as follows. Using the fact that every nonzero polynomial is a product of linear polynomials and polynomials of the form $(xa)^2+b$, where $b>0$, we see that for every $f(x)/g(x)\in F$, there are $a\in A$ and $b\in B$ such that $f$ and $g$ have constant sign on $(a,b)_R$; we define the sign of $f(x)/g(x)$ to be the sign it assumes on $(a,b)_R$. This makes $F$ an ordered field. Let $S$ be its real closure. For a given $\alpha\in S$, there exists $c\in R$ such that $\alpha\sim c$ whenever:
Thus $S$ is a proper rcf extensions of $R$ with the same Archimedean classes as $R$. 

