This is a very interesting question!

One way to interpret the question is like this: we have the structure
$\langle\mathbb{R},{+},{\cdot},{\lt}\rangle$, which is a [real-closed field](http://en.wikipedia.org/wiki/Real_closed_field), and Tarski proved
that this theory is both complete and decidable. You want to know,
what are the real-closed subfields of
$\langle\mathbb{R},{+},{\cdot},{\lt}\rangle$ that are decidable
with respect to the various models of computability on the reals?
Or at least, what are the order-types of these structures?

Note first that there are an enormous number of distinct
real-closed subfields of $\mathbb{R}$, namely, $2^{2^{\aleph_0}}$
many, since the transcendence degree of $\mathbb{R}$ over
$\mathbb{Q}$ is the continuum $2^{\aleph_0}$, and if one fixes a set $T$ of
continuum many algebraically independent reals, then for any
subset $A\subset T$, one may form $\mathbb{R}_A$, the
real-algebraic closure of $A$ in $\mathbb{R}$, and these will be
distinct real-closed subfields of $\mathbb{R}$, since they fill
different cuts in $\mathbb{Q}$. There are $2^{2^{\aleph_0}}$ many such subsets $A$. 
(I am unsure whether these
subfields all also have distinct order-types, which you asked
about; but it would seem very reasonable that they do---perhaps
someone can post about this.) 

It follows as a general conclusion that most real-closed subfields
of $\mathbb{R}$ are not computable with respect to any of the
usual concepts of computability on the reals, for there are simply
too many of them, and only countably many programs. Even if one
allows real parameters to be used as oracles in the computations,
this would still make only continuum many program/parameter
combinations, but strictly greater than the continuum many
real-closed subfields, and so not all of them are computable.

For some of the models, one can say more. In the
[Blum-Shub-Smale](http://en.wikipedia.org/wiki/Blum_Shub_Smale) model of computability, which you mentioned, for example,
in fact the only BSS-decidable real-closed subfield of $\mathbb{R}$ is the whole
field. The reason is that in the BSS model, no decidable set is
both dense and co-dense (see Corollary 1 of [Wesley Calvert, Ken
Kramer and Russell Miller, Noncomputable
Functions in the Blum-Shub-Smale Model](http://qcpages.qc.cuny.edu/~rmiller/BSSabstract.pdf)), but any proper subfield of $\mathbb{R}$ would be dense and co-dense.

Meanwhile, with the [infinite-time Turing machine]() model of
computability, there are infinitely many distinct real-closed
subfields of $\mathbb{R}$. The reason is that there are infinitely
many algebraically independent but ITTM computable reals, and
these will generate distinct ITTM computable real-closed
subfields.

In my article [Infinite time computable model theory](http://jdh.hamkins.org/infinitetimecomputablemodeltheory/), written
with Russell Miller, Dan Seabold and Steve Warner, we discuss in
section 2 various natural subfields with respect to ITTM
computability, including the fact that
$$\mathbb{R}_f\prec\mathbb{R}_w\prec\mathbb{R}_e\prec\mathbb{R}_a\prec\mathbb{R},$$
where these are, respectively, the finite-time decidable reals,
the infinite-time writable reals, the infinite-time eventually
writable reals, the infinite-time accidentally writable reals, and
all reals.

Another way to interpret the question, a way for which I have little to say, would be to ask not about real-closed subfields of $\mathbb{R}$, which are all Archimedean, but rather to ask which real-closed non-Archimedean fields have computable presentations (computable with respect to these real computational models). This interpretation might be a closer match with the Tennenbaum phenomenon in arithemtic.