This is an old question now but I had cause to look at it recently. I thought it was worthwhile pointing out that Carter's proof about the existence of nondegenerate maximal tori in Proposition 3.6.6 is something of a red herring. In fact, I claim that if $F$ is a Frobenius endomorphism (not generalised, so ruling out Suzuki and Ree groups) and $G^F$ is not of type ${}^2\mathrm{A}_{2n}$ then all $F$-stable maximal tori $T \leqslant G$ are non-degenerate if $q > 3$. The point about these assumptions is that I want any root to be orthogonal to any other root in its $F$-orbit. Note that one can also easily show the statement still holds for the groups $\mathrm{SU}_{2n+1}(q)$ by working in a matrix representation of the group. The proof below, however, will not work in this case.

Once one has this statement one can see that a non-degenerate maximal torus can have no rational regular elements, as pointed out by Jim.

To show that $T$ is non-degenerate we must show that for any root $\alpha \in \Phi \subseteq X(T)$ there exists an element $t \in T^F$ such that $\alpha(t) \neq 1$. We will denote by $\langle -,-\rangle : X(T) \times \check{X}(T)$ the usual perfect pairing between the character and cocharacter groups of $T$. Now as $F$ is a Frobenius endomorphism (not generalised) we have $F$ acts as $q\tau$ on the character group $X(T)$ respectively $q\check{\tau}$ on the cocharacter group $\check{X}(T)$. Here $\tau : X(T) \to X(T)$ and $\check{\tau} : \check{X}(T) \to \check{X}(T)$ are finite order automorphisms, of $\mathbb{Z}$-modules, such that $\tau(\Phi) = \Phi$ and $\check{\tau}(\check{\Phi}) = \check{\Phi}$.

Given $\alpha \in \Phi$ we denote by $k\geqslant 1$ the smallest integer such that $\check{\tau}^k(\check{\alpha}) = \check{\alpha}$. Given an element $\zeta \in \mathbb{F}_{q^k}^{\times}$ we then define an element $t_{\alpha}(\zeta) \in T^F$ by setting

\begin{equation*}
t_{\alpha}(\zeta) = \check{\alpha}(\zeta)\cdot\check{\tau}(\check{\alpha})(\zeta^q) \cdots \check{\tau}^{k-1}(\check{\alpha})(\zeta^{q^{k-1}})
\end{equation*}

Note this is clearly fixed by $F$ because for any element $\check{\beta}(\lambda) \in T$ we have $F(\check{\beta}(\lambda)) = \check{\tau}(\beta)(\lambda^q)$.

Now we can fix a $W = N_G(T)/T$-invariant bilinear form $(-,-)$ on the real vector space $V = \mathbb{R} \otimes_{\mathbb{Z}} X(T)$ such that

\begin{equation*}
\langle x, \check{\alpha}\rangle = \frac{2(x,\alpha)}{(\alpha,\alpha)}
\end{equation*}

for all $x \in V$ and $\alpha \in \Phi$, see the proof of 7.1.8 in Springer's LAGs. By 10.3.2(iii) of Springer's LAGs, and our assumption that $G^F$ is not of type ${}^2\mathrm{A}_{2n}$, we have for any $1 < i < k$ that

\begin{equation*}
0 = (\tau^i(\alpha),\alpha) \Rightarrow 0 = \langle \tau^i(\alpha),\check{\alpha}\rangle = \langle \alpha,\check{\tau}^i(\check{\alpha})\rangle.
\end{equation*}

Here I've just used the compatibility of the perfect pairing with $F$. Note this statement obviously fails in the case where $G^F$ is of type ${}^2\mathrm{A}_{2n}$ and also when the automorphism is a Suzuki/Ree automorphism. With this we see that

\begin{equation*}
\alpha(t_{\alpha}(\zeta)) = \zeta^{\langle \alpha,\check{\alpha}\rangle}\zeta^{q\langle \alpha,\check{\tau}(\check{\alpha})\rangle}\cdots \zeta^{q^{k-1}\langle \alpha,\check{\tau}^{k-1}(\check{\alpha})\rangle} = \zeta^{\langle \alpha,\check{\alpha}\rangle} = \zeta^2.
\end{equation*}

Hence, if there exists an element $\zeta$ such that $\zeta^2 \neq 1$ then $\alpha(t_{\alpha}(\zeta)) \neq 1$. In particular, if $q>3$ then the torus is non-degenerate.

In fact even when $q = 2$ one can have a non-degenerate maximal torus. By Proposition 3.6.7 of Carter’s book we must have $F$ is twisted. However one can easily check that $\mathrm{SU}_n(2)$, with $n \geqslant 3$, has a non-degenerate maximal torus.