Let $p$ be a prime. Let $B_{p,\infty}$ be a (unique) quaternion algebra ramified at exactly at $p$ and $\infty$ with a standard basis $1,i,j, k=ij=-ji$. Let $K=\mathbb{Q}(i) \subseteq B_{p,\infty}$ be a subalgebra with the fundamental discriminant $d_K$. Let $\omega= (1+i)/2$ if $d_K \equiv 1 \pmod 4$ or $\omega = i$ if $d_K\equiv 0 \pmod 4$ so that $O_K=\mathbb{Z}[\omega]$ is the maximal order in $K$. Let $O$ be a maximal order in $B_{p,\infty}$ and $\ell \neq p$ be a prime.
if $\omega \in O \otimes \mathbb{Z}_\ell$, then $O$ contains a quadratic order $\mathbb{Z}[c\omega]$ such that $\ell \nmid c$, where $c$ is the conductor.
The converse is obvious, but I'm not sure how to rigorously check if this is true. It's easier if I ask a similar question in one dimentional setting (like inside $K$).
Yes, this is a general fact that doesn't have to do with the fact that the quaternion algebra is ramified only at $p$--but I'm sure you have your reasons for that hypothesis! :) (I bet there's a proof using supersingular elliptic curves, but it's good to know we can do this using algebra, and this argument generalizes...)
I also had questions about your notation, but here's a thing that is separate from that which hopefully you can apply.
Let $B$ be a quaternion algebra over $\mathbb{Q}$ and let $\mathcal{O} \subset B$ be a maximal order. Then $\mathcal{O}$ is locally norm maximal (Example 28.5.16), so its associated class group $\mathrm{Cl}_{G(\mathcal{O})} \mathbb{Z}$, a quotient of the narrow class group of the integers, is trivial.
Let $q$ be a prime number which is unramified in $B$. Then $B$ is $\{q\}$-indefinite (Definition 28.5.1), that's just terminology. We are going to work over the global ring $R=\mathbb{Z}[1/q]$ where we allow denominators only at $q$, giving us the order $\mathcal{O}[1/q]$. We will conclude that we get an embedded order whose conductor is a power of $q$. Taking $q \neq \ell$ gives the result you are after. (And also hopefully if this all makes sense, you'll make acquaintance with several other important ingredients that you can make use of later!)
The first piece of content is strong approximation (Main Theorem 28.5.3), in particular its consequence in Theorem 28.5.5: the class set of $\mathcal{O}[1/q]$ as a $\mathbb{Z}[1/q]$-order is trivial: it maps bijectively to a trivial class group as noted above.
Now let $K$ be a quadratic field which embeds in $B$ and let $S=\mathbb{Z}_K$ be its ring of integers. We need the trace formula, Theorem 30.4.7. The summation on the left is just the one element; and this says that the number of embeddings of $S[1/q]$ into $\mathcal{O}[1/q]$ up to conjugation by $\mathcal{O}[1/q]^\times$ is a class number times a product of local embedding numbers, all of which are $\geq 1$ by Proposition 30.5.3.
If you're with me so far, then the thing we've proven is that $S[1/q]$ embeds in $\mathcal{O}[1/q]$. If $S=\mathbb{Z}[\gamma]$, then $q^e \gamma \in \mathcal{O}$ for some $e \geq 0$, and this gives the desired embedding.
We can relax substantially the conditions on $\mathcal{O}$, but this probably is already a lot to take at once! But hopefully it makes sense.