First, some generalities from algebraic topology (you can find these in any algebraic topology textbook). Suppose that $M$ is, say, a connected manifold (this works for a much larger class of topological spaces but we will not need this). Then the covering spaces of $M$ are classified (up to an isomorphism) by subgroups of the fundamental group of $M$. Furthermore, the degree of a covering map $M_H\to M$ corresponding to a subgroup $H\le \pi_1(M)$ equals the index of $H$ in $\pi_1(M)$. In particular, if $H$ is the trivial subgroup, $M_H\to M$ is the universal covering map and its degree equals the order (aka the cardinality) of $\pi_1(M)$. Thus, you are effectively asking about the order of the fundamental group $\pi_1(SO^+(p,q))$, where $p>1, q>1$.
Let $G$ be a connected Lie group and $K< G$ be its maximal compact subgroup. It is a general fact $G$ is diffeomorphic to the product of $K$ and some Euclidean space (see the references given here). (All maximal compact subgroups are conjugate to each other, hence, it does not matter which one you take.) In particular, $\pi_1(G)$ is isomorphic to $\pi_1(K)$. Thus, in order to compute $\pi_1(SO^+(p,q))$ we need to identify the/a maximal compact subgroup of $SO^+(p,q)$. You can find this, for instance, in
Helgason, Sigurdur, Differential geometry, Lie groups, and symmetric spaces., Graduate Studies in Mathematics. 34. Providence, RI: American Mathematical Society (AMS). xxvi, 641 p. (2001). ZBL0993.53002.
A free (and better, in my opinion) reference is Brian Konrad's online notes "Examples of maximal compact subgroups".
In any case, the maximal compact subgroup of $O(p,q)$ is $O(p)\times O(q)$, which yields $SO(p)\times SO(q)$ for $SO^+(p,q)$. Thus,
$$
\pi_1(SO^+(p,q))\cong \pi_1(SO(p))\times \pi_1(SO(q)).
$$
You can find a computation of $\pi_1(SO(n)), n\ge 2$ in many places, for instance, in the answers to this question. But you probably already know that $\pi_1(SO(2))\cong {\mathbb Z}$ and $\pi_1(SO(n))\cong {\mathbb Z}_2$, $n\ge 3$.
Thus, we obtain:
(a) If $p=q=2$, then $\pi_1(SO(p))\times \pi_1(SO(q))\cong {\mathbb Z}^2$ (hence, is infinite).
(b) If $p>2, q=2$, then $\pi_1(SO(p))\times \pi_1(SO(q))\cong {\mathbb Z}_2\times {\mathbb Z}$ (hence, is infinite).
(c) if $p>2, q>2$, then $\pi_1(SO(p))\times \pi_1(SO(q))\cong {\mathbb Z}_2\times {\mathbb Z}_2$ has order four.
In particular, in all cases, $\pi_1(SO^+(p,q))$ has order 4 or is infinite; it never has order two as long as $p>1, q>1$.
The bottom line is that for $p>1, q>1$ the universal cover of $SO^+(p,q)$ is never the (2-fold) spin-cover $Spin(p,q)$. I am not sure about the origin of the claim in the physics literature that $Spin(p,q)$ is the universal cover of $SO^+(p,q)$. Most likely, somebody have read that the universal cover of $SO^+(n,1), n\ge 3$, is $Spin(n,1)$ (which is true, of course) and extrapolated (without thinking) to all indefinite orthogonal groups.
