Over at math.SE, I give two polynomials $F(x,y,z)$, $G(x,y,z)$, whose common zero locus is the trefoil knot, and which are smooth and transverse there. Therefore, $(\nabla F) \times (\nabla G)$ is a vector field tangent to the knot.
I think you should be able to get a shorter answer as follows. Take the sphere $S$ given by $x_1^2+y_1^2+x_2^2+y_2^2=1$ in $\mathbb{R}^4$. The vector field $$p(-y_1 \frac{\partial}{\partial x_1}+x_1 \frac{\partial}{\partial y_1})+q(-y_2 \frac{\partial}{\partial x_2}+x_2 \frac{\partial}{\partial y_2})$$ is tangent to $S$, and its flow lines are of the form $$(x_1, y_1, x_2, y_2) = (r_1 \cos(p t), r_1 \sin(p t), r_2 \cos(q t), r_2 \sin(q t))$$ which is a $(p,q)$ torus knot in $S$.
Now make a stereographic change of coordinates to get a vector field on $\mathbb{R}^3$ rather than $S^2$. That will give you a vector field with rational functions as coefficients, and clearing out the denominators will just change the speed of the flow.
But I don't feel up to doing that computation right now, so I'll leave the details to you.