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 any polynomial over the positive integers evaluated at $a$ is still less than $c$ in $M$. So the $\mod c$ part of $F$ never arises when evaluating a polynomial over $\mathbb{N}$ at $a$. So the only way such two such polynomials can agree at $a$ is if they are the same. Thus, no polynomial over $\mathbb{Z}$ at $a$ can be $0$ except for the zero polynomial, and so $a$ is transcendental over $\mathbb{Q}$ in $F$.