Here's a more naive approach than that suggested by Jim in his answer, and Kreck in comments.

One can explicitly write down the Sylow $p$-subgroup as a matrix group using a basis of hyperbolic pairs. If you order this basis correctly, then there will be a Sylow $p$-subgroup of the associated classical group which will be a bunch of upper-triangular matrices. (See Kleidman & Liebeck *"The subgroup structure of the finite classical groups"* for a full description.)

(Note: I'm avoiding questions about which version of the group you are interested in. Barring some pathological cases, the natural matrix group version of the classical group is the universal version. In any case, if you avoid bad primes, as Jim describes above, the exponent of a Sylow $p$-subgroup of your group won't depend on which version you're using.)

So, for example, if I take $e_1,\dots, e_k, f_1,\dots, f_k$ to be such a basis for a vector space $V$ over $\mathbb{F}_q$ where $q$ is odd, and I assume that $V$ is equipped with a non-degenerate alternating bilinear form, then the isometry group of the form will be the symplectic group ${\mathrm Sp}_{2k}(q)$ and one can choose a Sylow $p$-subgroup of $G$ such that all elements are upper-triangular. Once you've done this it's easy to observe that $G$ contains an element
$$\left(\begin{array}{ccccc}
1 & 1 & & & \\
& 1 & 1 & & \\
& & \ddots \ddots & & \\
& & & 1 & -1 \\
& & & & 1
\end{array}\right).$$
(On my super-diagonal I have $k$ lots of $1$ and $k-1$ lots of $-1$.)
I haven't double-checked, but I believe this is an element of maximal exponent. Its order is the least power of $p$ greater than $k+1$.

It's the same story with orthogonal groups. Again I haven't double-checked that I really am obtaining an element of maximal order, but this should be straight-forward.

In any case, these calculations suggest the following result. 

>**Conjecture**: Let $P$ be a Sylow $p$-subgroup of an untwisted classical group $G$ over $\mathbb{F}_q$, and suppose that $p$ is not a bad prime. Then the exponent of $p$ is equal to the least power of $p$ greater than ${\mathrm rk}(G)+1$.

(I'm stating the conjecture conservatively. It's possible that it holds for twisted groups, and for bad primes, but that would just be idle speculation!)