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 $M$. It follows that all the standard
powers of $a$ are still bounded by $c$, even if you also
multiply by rational numbers. So the $\mod c$ part of $F$
never arises when evaluating a polynomial over $\mathbb{Q}$
at $a$. So the only way such a polynomial can be zero at $a$ is if it is the zero polynomial, and so $a$ is transcendental over $\mathbb{Q}$.