The multiplicative group $\mathbb{F}_{q^2}^\times$ acts by multiplication on $\mathbb{F}_{q^2}$ which is $2$D as a vector space over $\mathbb{F}_q$. Thus, each element $z\in\mathbb{F}_{q^2}^\times$ is represented by a $2\times 2$ matrix $L_z$ over $\mathbb{F}_q$ (by choosing a basis). In this case, since $q\equiv3$ mod $4$ we can say $\mathbb{F}_{q^2}^\times$ has an element $i$ of order $4$, but $\mathbb{F}_q^\times$ doesn't, so $\mathbb{F}_{q^2}=\mathbb{F}_q(i)$ and every element of $\mathbb{F}_{q^2}$ is expressible as a "complex number" $z=x+yi$ with $x,y\in\mathbb{F}_q$, with corresponding matrix $L_z=(\begin{smallmatrix} x & -y\, \\ y & ~x \end{smallmatrix})$. The determinant is
$$\det L_z=N_{\small\mathbb{F}_{{\large q}^2}\mid\mathbb{F}_{\large q}}(z)=z\tau(z)=z^{q+1}, $$
with the Frobenius automorphism $\tau\in\mathrm{Gal}(\mathbb{F}_{q^2}\!\mid\!\mathbb{F}_q)$ given by $\tau(z)=z^q$, or with respect to our basis $\tau(x+yi)=x-yi$ ("complex conjugation").
Thus, the $L_z\in\mathrm{SL}_2\mathbb{F}_q$ correspond to $(q+1)$st roots of unity $z$. Indeed, $\mathbb{F}_{q^2}^{\times}$ is cyclic, so its $2$-Sylow subgroup is cyclic as well, say generated by a root $\omega\in\mathbb{F}_{q^2}^\times$ corresponding to $\varpi=L_\omega$. Since the cardinality $|\mathrm{SL}_2\mathbb{F}_q|$ is $(q-1)q(q+1)$, we can say $\langle\varpi\rangle$ has index $2$ (the largest power of $2$ dividing $q-1$) within some $2$-Sylow of $\mathrm{SL}_2\mathbb{F}_q$. Thus, we need find only one more element $\sigma\in\mathrm{SL}_2\mathbb{F}_q$ to create a full $2$-Sylow $\langle\varpi,\sigma\rangle$.
Denote $\iota=L_i=(\begin{smallmatrix}0 & -1\, \\ 1 & ~0\end{smallmatrix})$. Ideally, we can choose $\sigma$ so that $\langle\sigma,\iota\rangle\cong Q_8$. Setting $\sigma=(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix})$, the condition $\iota\sigma=-\sigma \iota$ is equivalent to $d=-a$ and $c=b$ so $\sigma=(\begin{smallmatrix} a & ~b \\ b & -a\,\end{smallmatrix})$. The conditions $\det\sigma=1$ and $\sigma^2=I_2$ are both equivalent to $a^2+b^2=-1$. This is guaranteed to have a solution in $\mathbb{F}_p$ by a standard argument: the number of squares $a^2$ is $(p+1)/2$, which is also the number of elements of the form $(-1-b^2)$s, so these two sets overlap.
$\big[$ Fun exercise: show $\sigma=(\begin{smallmatrix} a & ~b \\ b & -a\, \end{smallmatrix})$ and $\rho=(\begin{smallmatrix} c & ~d \\ d & -c\, \end{smallmatrix})$ represent the same coset of $\langle\varpi\rangle$ for two different solutions $(a,b)$ and $(c,d)$; IOW, that $\rho^{-1}\sigma$ is expressible as $L_z$ for some $z\in\mathbb{F}_{q^2}^\times$. $\big]$
Writing $\omega=u+vi$, observe $\omega\overline{\omega}=\omega\omega^q=1$ so $\overline{\omega}=\omega^{-1}$, and so we may verify
$$ \sigma \varpi \sigma^{-1} \,=\,
\sigma L_{u+vi} \sigma^{-1} \,=\,
\sigma(u I_2+ v \iota)\sigma^{-1} \,=\,
u I_2 - v \iota \,=\,
L_{u-vi} = \varpi^{-1}. $$
In conclusion, $\langle \varpi,\sigma\rangle$ is generalized quaternion and is a $2$-Sylow subgroup of $\mathrm{SL}_2\mathbb{F}_q$.
It is probably possible to frame this in terms of split quaternions, since the split quaternions over a field $\mathbb{F}$ are isomorphic to the $2\times2$ matrix algebra $M_2\mathbb{F}$ as an $\mathbb{F}$-algebra.
The $p$-Sylow subgroups of $\mathrm{GL}_n\mathbb{F}_q$ can be described as follows. Note its order is
$$ |\mathrm{GL}_n\mathbb{F}_q|=q^{\binom{n}{2}}(q^n-1)(q^{n-1}-1)\cdots(q-1). $$
If $q$ is a power of $p$ then the unitriangular matrices form a $p$-Sylow.
Otherwise, let $e=\mathrm{ord}_pq$ be the multiplicative order of $q$ mod $p$ (or mod $4$ if $p=2$), i.e. $e$ is the minimal exponent for which $p\mid(q^e-1)$. Picking a basis for $\mathbb{F}_{q^e}$, every scalar $z$ can be turned into a matrix $L_z$, and the Galois group $\mathrm{Gal}(\mathbb{F}_{q^e}\!\mid\!\mathbb{F}_q)$ also acts linearly on $\mathbb{F}_{q^e}$ as a $\mathbb{F}_q$-vector space. This gives us an embedding $\mathbb{F}_{q^e}^\times\rtimes\mathrm{Gal}(\mathbb{F}_{q^e}\!\mid\!\mathbb{F}_q)\hookrightarrow \mathrm{GL}_e\mathbb{F}_q$.
Treating these as $e\times e$ block matrices, we can put $\lfloor n/e\rfloor$ of these blocks diagonally in an $n\times n$ matrix (padding with an identity matrix in the lower right corner as necessary). If we apply block-permutation matrices representing $S_{\lfloor n/e\rfloor}$ we get "block monomial" matrices, giving an embedding
$$ \big(\mathbb{F}_{q^e}^\times \rtimes \mathrm{Gal}(\mathbb{F}_{q^e}\!\mid\!\mathbb{F}_q)\big)\wr S_{\lfloor n/e\rfloor} \hookrightarrow \mathrm{GL}_n\mathbb{F}_q $$
If we replace each of $\mathbb{F}_{q^e}^\times$, $\mathrm{Gal}(\mathbb{F}_{q^e}\!\mid\!\mathbb{F}_q)$, $S_{\lfloor n/e\rfloor}$ with one of their own $p$-Sylow subgroups, the product above is still well-defined and restricting the embedding gives a $p$-Sylow of $\mathrm{GL}_n\mathbb{F}_q$. Note the first two of these groups is cyclic, and the $p$-Sylow of symmetric groups I described in here. Indeed, generally $e<p$ so $\mathrm{Gal}(\mathbb{F}_{q^e}\!\mid\!\mathbb{F}_q)$ has trivial $p$-Sylow, unless $p=2$ and $q\equiv3\bmod 4$ in which case the $2$-Sylow is $\cong\mathbb{Z}_2$ ("complex conjugation").
$\big[$ Fun exercise: Show that this construction of the $2$-Sylow of $\mathrm{GL}_2\mathbb{F}_q$, when intersected with $\mathrm{SL}_2\mathbb{F}_q$, leads to the original $2$-Sylow of $\mathrm{SL}_2\mathbb{F}_q$ I described above. $\big]$
