As shown by David, the answer is no for line bundles.

More generally, the answer in no even for higher dimensional vector bundles. In fact, in his paper "Vector bundle over ruled surfaces birationally equivalent to ruled surfaces" Murthy proves the following results:

(1) Let $V$ be an irreducible affine non-singular surface defined over an algebraically closed field $k$ and such that $V$ is birational to $C \times \mathbb{P}^1$, vhere $C$ is a curve. Then any vector bundle over $V$ is a direct sum of a trivial bundle and a line bundle. In general, there are infinitely many equivalence classes of line bundles (as shown by David for $V= \mathbb{P}^1 \times \mathbb{P}^1 \setminus \Delta$).

(2) There exist an affine, nonsingular rational variety of dimension $3$ over which there are infinitely many non-isomorphic indecomposable vector bundles of rank $>1$.

As shown by David, the answer is no for line bundles.

More generally, the answer in no even for higher dimensional vector bundles. In fact, in his paper "Vector bundle over affine surfaces birationally equivalent to ruled surfaces" Murthy proves the following results:

(1) Let $V$ be an irreducible affine non-singular surface defined over an algebraically closed field $k$ and such that $V$ is birational to $C \times \mathbb{P}^1$, vhere $C$ is a curve. Then any vector bundle over $V$ is a direct sum of a trivial bundle and a line bundle. In general, there are infinitely many equivalence classes of line bundles (as shown by David for $V= \mathbb{P}^1 \times \mathbb{P}^1 \setminus \Delta$).

(2) There exist an affine, nonsingular rational variety of dimension $3$ over which there are infinitely many non-isomorphic indecomposable vector bundles of rank $>1$.