The implication does not hold.

Let $K$ be the real numbers, $F$ the complex numbers, $A=K[X,Y,Z]/(X^2+Y^2+Z^2-1)$, and $M$ the kernel of the map $A^3\rightarrow A$ given by the unimodular row $(X,Y,Z)$.

Then $M$ cannot be free, by the same argument I gave in my answer to this question.

But $M$ becomes free after tensoring with the complex numbers. To see this, write $U=(X+iY)/2$, $V=(X-iY)/2$. Then it suffices to show that $(X,Y,Z)=(U+V,-iU+iV,Z)$ can be transformed via elementary operations to $(1,0,0)$ (so that its kernel is isomorphic to the kernel of $(1,0,0)$, which is evidently free).

To construct such a series of elementary transformations, first transform $(U+V,-iU+iV,Z)\rightarrow (2U,-iU+iV,Z)\rightarrow (2U,iV,Z)$ and then note that $2U$ is equal to 1 mod $(iV,Z)$.

The implication does not hold.

Let $K$ be the real numbers, $F$ the complex numbers, $A=K[X,Y,Z]/(X^2+Y^2+Z^2-1)$, and $M$ the kernel of the map $A^3\rightarrow A$ given by the unimodular row $(X,Y,Z)$.

Then $M$ cannot be free, by the same argument I gave in my answer to this question.

But $M$ becomes free after tensoring with the complex numbers. To see this, write $U=(X+iY)/2$, $V=(X-iY)/2$. Then it suffices to show that $(X,Y,Z)=(U+V,-iU+iV,Z)$ can be transformed via elementary operations to $(1,0,0)$ (so that its kernel is isomorphic to the kernel of $(1,0,0)$, which is evidently free).

To construct such a series of elementary transformations, first transform $(U+V,-iU+iV,Z)\rightarrow (2U,-iU+iV,Z)\rightarrow (2U,iV,Z)$ and then note that $2U$ is equal to 1 mod $(iV,Z)$.