Personally, I think this problem is ill-posed.  What geometric properties would the OP like to assume about $S$?  What does "constant" mean -- constant on the set of rational points, or truly a constant polynomial?  Also, what precisely does $d$ "significantly smaller" than $q$ mean?  There are "surfaces" in $\mathbb{F}_q^3$ that have few rational points, e.g., the vanishing set of $x^2-x$ is a surface with only $q^2$ points.  Worse yet, a polynomial like $P(x,y,z) = [(x-1)^2-1]^2 = x^2(x-2)^2$ will vanish on all tangent vectors at these $q^2$ points.  Of course $S$ is reducible, but the OP says nothing about irreducibility.  I suspect that there are similar irreducible examples: the crucial point is that the (few) rational points (mostly) lie on a small number of curves cut out by low degree equations.

<B>Edit.</B> Vivek complains that my example $S$ above is disconnected.  However, one can use my second suggestion to easily produce connected counterexamples.  Start with a disconnected set of small size compared to $q$, e.g., $\{(0,0,0),(1,0,0)\}$, and a small number $d$ of low degree defining equations, e.g., $x^2-x=y=z=0$ and $d=3$.  Now take a normic form $g$ over $\mathbb{F}_q$ of degree $d$ in $d$ variables: these always exist over finite fields (cf. Lang's thesis). Now plug in the defining equations of your disconnected set for the variables of the normic form to get a new polynomial $h$, e.g., $h(x,y,z) = g(x^2-x,y,z)$.  Now the only rational points of the zero set of $h$ will be the points in the original disconnected set, e.g., $\{(0,0,0),(1,0,0)\}$.  Now you can use a polynomial $P$ such as my polynomial above.  Really this has <B>NOTHING</B> to do with $S$, and only to do with the (incredibly small) set of rational points of $S$.