First, you haven't actually specified a particular field,
since the field $F$ that you have will depend on your
choice of $c$ and of $M$. For example, different
nonstandard models can seriously affect even the
cardinality of the field $F$ that you produce, so they are
not all the same. (A Lowenheim-Skolem argument shows that
$F$ can be found as you describe of any desired infinite
cardinality.)

But to answer your question, none of these fields is
algebraic over $\mathbb{Q}$. To see this, let $a$ be any
nonstandard integer in $M$ whose finite powers are bounded
below $c$ in $M$ (see below). It follows that any
polynomial over $\mathbb{N}$ evaluated at $a$ is still less
than $c$ in $M$. So the $\mod c$ part of the field
operations of $F$ never arise when evaluating a polynomial
over $\mathbb{N}$ at $a$. Thus, the problem reduces to
showing that if $p$ is a nontrivial polynomial over
$\mathbb{Z}$, then $p(a)\neq 0$ for nonstandard $a$ in $M$,
and this follows because the basic eventually-unbounded
asymptotic behavior of such polynomials is provable in your
theory. Thus, $a$ is transcendental over $\mathbb{Q}$ in
your field $F$.

<b>Edit.</b> Finally, here is a quick-and-dirty way to see
that there is such a nonstandard element $a$, whose finite
powers are bounded by $c$ in $M$. Let $a$ be the nearest
nonstandard integer to $c^{1/N}$, where $N=\sqrt{\log c}$
as interpreted discretely in $M$. Since $N$ is nonstandard,
it follows that the finite powers of $a$ are below $c$, and
since $\log a\equiv\frac 1N\log c$, it follows by the choice of
$N$ that $a$ is nonstandard. But I expect that there is an
easier method.