(I wrote this earlier and didn't finish posting it. It has some overlap with Remy's answer, but I figured it was worth posting to provide a slightly different perspective.)

No. For example the elliptic curves $y^2 = x^3 +x+1$ and $y= x^3 + x-1$ are isomorphic over $\mathbb C$ by $ x \mapsto -x, y \mapsto iy $ but only isomorphic over $\mathbb F_q$ if $q \equiv 1 \bmod 4$ so that a square root of $-1$ lives in $\mathbb F_q$, i.e. not just over fields of sufficiently large characteristic. This is just because elliptic curves are only isomorphic if the Weierstrass equations are related by a linear change of coordinates, and if the coefficient of $x^2=0$ this can only be scaling the coordinates, and scaling $x$ by $+1$ is the only thing that works which requires us to scale $y$ by a square root of $-1$.

These elliptic curves are also not isomorphic in the Grothendieck ring. They don't even have the same number of $\mathbb F_q$-points (for most $q$ not congruent to $1$ mod $4$). This is because the quadratic twist has the function of replacing each $x$ coordinate with two solutions $y$ by one with no solutions, and vice versa, meaning if one curve has $n$ points the other has $2(q+1)-n$ points, so they only have the same number of points if $q+1=n$, which rarely holds.

The problem here is that it's easy to produce situations where there is enough rigidity that there is an isomorphism over $\mathbb F_q$ if and only if the isomorphism over $\mathbb C$ becomes an isomorphism over $\mathbb F_q$, but if the field of definition of the isomorphism is a number field $K$, the isomorphism only extends to $\mathbb F_q$ if $\mathbb F_q$ contains one of the residue fields of $K$.

On the other hand, your last bullet point is correct (ignoring the "in the affirmative case"). One doesn't even need the "for sufficiently large characteristic" as it is included in the "extends $R$" (one can just invert small primes in $R$ to ensure only large characteristic finite fields can extend it). For an affine or projective variety, it's easy to check that an isomorphism is given by finitely many parameters (the coefficients of image under the isomorphism of the generators of one coordinate ring as polynomias in the generators of the other coordinate ring, and vice versa for the inverse map). Let $R_0$ be the ring generated by these parameters. Then there is an isomorphism between the varieties defined over $R_0$ (since the two compositions being the identities is a closed condition). Then the isomorphism also lives over each finite field that extends $R_0$.

(I wrote this earlier and didn't finish posting it. It has some overlap with Remy's answer, but I figured it was worth posting to provide a slightly different perspective.)

No. For example the elliptic curves $y^2 = x^3 +x+1$ and $y^2= x^3 + x-1$ are isomorphic over $\mathbb C$ by $ x \mapsto -x, y \mapsto iy $ but only isomorphic over $\mathbb F_q$ if $q \equiv 1 \bmod 4$ so that a square root of $-1$ lives in $\mathbb F_q$, i.e. not just over fields of sufficiently large characteristic. This is just because elliptic curves are only isomorphic if the Weierstrass equations are related by a linear change of coordinates, and if the coefficient of $x^2=0$ this can only be scaling the coordinates, and scaling $x$ by $+1$ is the only thing that works which requires us to scale $y$ by a square root of $-1$.

These elliptic curves are also not isomorphic in the Grothendieck ring. They don't even have the same number of $\mathbb F_q$-points (for most $q$ not congruent to $1$ mod $4$). This is because the quadratic twist has the function of replacing each $x$ coordinate with two solutions $y$ by one with no solutions, and vice versa, meaning if one curve has $n$ points the other has $2(q+1)-n$ points, so they only have the same number of points if $q+1=n$, which rarely holds.

The problem here is that it's easy to produce situations where there is enough rigidity that there is an isomorphism over $\mathbb F_q$ if and only if the isomorphism over $\mathbb C$ becomes an isomorphism over $\mathbb F_q$, but if the field of definition of the isomorphism is a number field $K$, the isomorphism only extends to $\mathbb F_q$ if $\mathbb F_q$ contains one of the residue fields of $K$.

On the other hand, your last bullet point is correct (ignoring the "in the affirmative case"). One doesn't even need the "for sufficiently large characteristic" as it is included in the "extends $R$" (one can just invert small primes in $R$ to ensure only large characteristic finite fields can extend it). For an affine or projective variety, it's easy to check that an isomorphism is given by finitely many parameters (the coefficients of image under the isomorphism of the generators of one coordinate ring as polynomias in the generators of the other coordinate ring, and vice versa for the inverse map). Let $R_0$ be the ring generated by these parameters. Then there is an isomorphism between the varieties defined over $R_0$ (since the two compositions being the identities is a closed condition). Then the isomorphism also lives over each finite field that extends $R_0$.