You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where
$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$
So that the vector field
$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$
is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:
$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$
this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is
$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$
In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, asymptotically rigidly with a constant period right near the z-axis.
I think you flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:
$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$
(this is different from the previous thing). That's why you had a difference of opinion.