This becomes clearer when considered in the broader context of special orthogonal groups of non-degenerate quadratic forms over general fields. (You are considering the case of the ground field $\mathbf{R}$, over which an $n$-dimensional non-degenerate quadratic space is characterized up to isomorphism by the signature $(p, q)$: integers $p, q \ge 0$ with $p+q=n$.)

In general, let $(V, Q)$ be a non-degenerate quadratic space of dimension $n \ge 1$ over a field $k$, and one can ask when the connected reductive $k$-group $G = {\rm{SO}}(Q)$ is quasi-split.

By non-degeneracy, if $v \in V - \{0\}$ is a non-trivial zero of $Q$ then there is a hyperbolic plane $H$ in $V$ containing $v$. If
$H$ is a hyperbolic plane contained in $V$ (i.e., the span of 2 nonzero $Q$-isotropic vectors $v_0, v'_0$ satisfying $B_Q(v_0,v'_0)=1$, where $B_Q(v,v')=Q(v+v')-Q(v)-Q(v')$ is the associated symmetric bilinear form) then $V$ is the orthogonal sum of $H$ and its $B_Q$-orthogonal space inside $V$.

By iterating this, we arrive at an orthogonal decomposition of quadratic spaces
$$(V, Q) \simeq H_1 \perp \dots \perp H_r \perp (V_0, Q_0)$$
where each $H_j$ is a hyperbolic plane (so $0 \le r \le [n/2]$) and $Q_0$ is non-degenerate and $k$-anisotropic (i.e., $Q_0$ has no nontrivial zeros in $V_0$). In other words, in suitable linear coordinates
$$Q = x_1 x_2 + \dots + x_{2r-1}x_{2r} + Q_0(x_{2r+1}, \dots, x_n)$$
with $k$-anisotropic $Q_0$. By Witt's Cancellation Theorem, $r$ is the maximal number of pairwise orthogonal hyperbolic planes contained in $V$.

I claim that $r$ is the $k$-rank of $G$; i.e., the common dimension of the maximal $k$-split tori of $G$. More specifically, let $S = \mathbf{G}_m^r \subset G$ be the $k$-split subtorus $\prod {\rm{SO}}(H_j)$. Concretely, the $j$th factor acts through $\mathbf{G}_m$-scaling on an isotropic basis $\{e_j, e'_j\}$ of $H_j$ via $t.e_j = te_j$ and $t.e'_j = t^{-1}e'_j$ and acts trivially on $V_0$ and every $H_i$ for $i \ne j$. Then it is clear from weight-space considerations for the $S$-action on $V$ that
$$Z_G(S) = S \times {\rm{SO}}(Q_0).$$
But it is a general (not difficult) fact that a non-degenerate quadratic space over $k$ (with any dimension $\ge 0$, such as $V_0$!) is $k$-anisotropic if and only if the associated special orthogonal group does not contain $\mathbf{G}_m$ as a $k$-subgroup. Hence, the $k$-anisotropicity of $Q_0$ implies that $Z_G(S)/S$ contains no nontrivial split tori, so $S$ is indeed maximal as a $k$-split torus in $G$.

In the structure theory for connected reductive groups over a field $k$, the Levi factors of the minimal parabolic $k$-subgroups are precisely the centralizers of the maximal $k$-split tori, and a parabolic $k$-subgroup is a Borel subgroup precisely when its Levi factors (or better: its quotient by its unipotent radical) is commutative. Hence, $G$ is quasi-split if and only if ${\rm{SO}}(Q_0)$ is commutative, which is to say $\dim V_0 \le 2$.

Returning now to the motivating situation over $\mathbf{R}$, the signature of $(V_0, Q_0)$ is clearly $(r+d_0,r)$ or $(r, r+d_0)$ where $d_0 = \dim V_0$ and the two options for the signature correspond respectively to the $\mathbf{R}$-anisotropic $(V_0, Q_0)$ being positive-definite or negative-definite. Hence, in terms of the classical notation ${\rm{SO}}(p,q)$ clearly $|p-q| = d_0$, so we have proved that $G$ is quasi-split if and only if $|p-q| \le 2$.