I'm not sure why you emphasize linear combinations with coefficients $\pm 1$ to the exclusion of other possible rational coefficients.  

In any case, the degree $d=[\mathbf Q(\sqrt{a_1},\ldots,\sqrt{a_n}):\mathbf Q]$ is the order of $\langle a_1,\ldots,a_n\rangle$ in $\mathbf Q^\times/(\mathbf Q^\times)^2$, so it is a power of $2$ that is basically counting how multiplicatively independent the $a_i$'s are modulo rational squares. An approach to proving that multiplicative independence of $a_i$'s mod squares implies their square roots are linearly independent over $\mathbf Q$ by using reduction mod $p$ for a large prime $p$ can be read [here][1].

(Update) In the ring $\mathbf Z[\sqrt{a_1},\ldots,\sqrt{a_n}]$, each nonzero prime ideal $\mathfrak p$ gives us a quotient ring $\mathbf Z[\sqrt{a_1},\ldots,\sqrt{a_n}]/\mathfrak p$, which is a *finite field* whose characteristic $p$ is the prime number such that (i) $\mathfrak p \cap \mathbf Z = p\mathbf Z$ or (ii) $(p)\subset\mathfrak p$ as ideals (these are equivalent properties). That finite field is called the residue field at $\mathfrak p$. Because $\sqrt{a_i}^2 = a_i$ in the ring, we have $\sqrt{a_i}^2 \equiv a_i \bmod \mathfrak p$ in the residue field. The residue field at $\mathfrak p$ is generated as a ring by all $\sqrt{a_i} \bmod \mathfrak p$, so this field is either $\mathbf F_p$ or $\mathbf F_{p^2}$, and it is $\mathbf F_p$ if and only if all $a_i \bmod p$ are squares in $\mathbf F_p$. The density of primes $p$ such that all $a_i \bmod p$ are squares in $\mathbf F_p$ is $1/d$ where $d$ is the degree $[\mathbf Q(\sqrt{a_1},\ldots,\sqrt{a_n}):\mathbf Q]$ I mentioned earlier (a certain power of $2$).

Any $\mathbf Z$-linear relation $c_1\sqrt{a_1}+\ldots+c_n\sqrt{a_n}=0$ in the ring leads to an $\mathbf F_p$-linear relation $c_1\sqrt{a_1}+\ldots+c_n\sqrt{a_n} \equiv 0 \bmod \mathfrak p$ in the residue field at $\mathfrak p$. This is a good way to carry out the process of turning a relation in characteristic $0$ into a relation in characteristic $p$. 

In fact, *all* ways of picking a square root of each $a_i \bmod p$ in $\mathbf F_{p^2}$ arise from some choice of prime ideal $\mathfrak p$ where $(p) \subset \mathfrak p$: if $\alpha_i^2 = a_i$ in $\mathbf F_{p^2}$, then there is a unique ring homomorphism 
$$
\mathbf Z[x_1,\ldots,x_n]\to \mathbf F_{p^2}
$$ 
where $x_i\mapsto \alpha_i$ for all $i$. This map kills $x_i^2-a_i$ for all $i$, so we get an induced ring homomorphism
$$
\mathbf Z[x_1,\ldots,x_n]/(x_1^2-a_1,\ldots,x_n^2-a_n)\to \mathbf F_{p^2},
$$
and the domain is isomorphic to $\mathbf Z[\sqrt{a_1},\ldots,\sqrt{a_n}]$ by identifying the coset of $x_i$ with $\sqrt{a_i}$, so we get a (unique) ring homomorphism
$$
\mathbf Z[\sqrt{a_1},\ldots,\sqrt{a_n}]\to \mathbf F_{p^2}
$$
where $\sqrt{a_i} \mapsto \alpha_i$ for all $i$. The kernel has to be a nonzero prime ideal, say $\mathfrak p$, so we get an induced ring embedding
$$
\mathbf Z[\sqrt{a_1},\ldots,\sqrt{a_n}]/\mathfrak p \hookrightarrow \mathbf F_{p^2}
$$
with image $\mathbf F_p$ or $\mathbf F_{p^2}$ in which $\sqrt{a_i} \bmod \mathfrak p =  \alpha_i$ for all $i$. This does not just preserve additive relations among all $\sqrt{a_i}$ when going from characteristic zero to characteristic $p$, but all multiplicative relations as well (it is a ring homomorphism).



[1]:https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/