Here is another example similar to Angelo's construction of a non-toral diagonalizable subgroup of a reductive group. I'll suppose that the characteristic is not 2.
Let $G = SO(V) = SO(V,\beta)$ for $\dim V > 2$, and write $V$ as an orthogonal sum
$V = U \perp W$ for $0 < \dim U < \dim V$ with $\dim U$ *even*,
such that the restriction of $\beta$ to $U$ and $W$ is non-degenerate.

Let $t \in G$ act as the identity on $W$ and as $-1$ on $U$. Then the
centralizer $M=C_G(t)$ identifies with the subgroup
{$(x,y) \in O(U) \times O(W) \mid \det(x) = \det(y)$}. In particular,
this centralizer is not connected: $M/M^0$ has order 2.

One can evidently choose an involution $s \in M \setminus M^0$, and then
$D = \langle t,s\rangle$ is a diag. subgroup of $G$ which is contained
in no maximal torus.

Part of this construction can be made in char. 2. Instead of $t$, you have
to take a non-smooth subgroup $\mu \simeq \mu_2$, essentially given by
the action of a semisimple element $X \in \operatorname{Lie}(G)$ ($X$ should
act as $1$ on $U$ and $0$ on $W$). Then $M=C_G(\mu) = C_G(X)$ is again
disconnected (well, now you can't argue by determinants) with component
group of order $2$. But this doesn't seem to lead to a non-toral diagonalizable
subgroup (any finite order element representating the non-trivial
coset of $M/M^0$ has a non-trivial unipotent part).

the cyclic case: If $D = \langle s \rangle$ is a cyclic diagonalizable subgroup of a connected linear algebraic group $G$, then $s$ is a semisimple element of $G$ (of finite order). In particular, $s$ and hence $D$ is contained in a maximal torus of $G$. Indeed, by [Borel LAG,11.10] $s$ is contained in a Borel subgroup of $G$, and then the claim follows from the connected solvable case [Borel LAG,10.6]. – George McNinch Apr 8 '11 at 13:15