The irreducibles of the Sylow 2-subgroups of ${\rm GL}(n,q)$, $q$ odd, have indeed trivial Schur indices: Let $S$ be such a Sylow 2-subgroup. First, observe that the natural module $(\mathbb{F}_q)^n$ splits into a sum of simple modules $U_1\oplus \dotsb \oplus U_l$, and the dimension of each simple module is a power of $2$. This shows that $S \cong S_1 \times \dotsb \times S_l $, where the $S_i$'s are Sylow 2-subgroups of a ${\rm GL}(2^k, q)$. Thus, w.l.o.g. we may assume that $n=2^k$. Then I use induction on $k$. First note that if $S$ is a Sylow 2-subgroup of ${\rm GL}(2^k, q)$, then $$ T=\lbrace \begin{pmatrix} s & \\ & t \end{pmatrix} \mid s, t\in S \rbrace \cup \lbrace \begin{pmatrix} & s \\ t & \end{pmatrix} \mid s,t\in S\rbrace \cong S\wr C_2 $$ is a Sylow 2-subgroup of ${\rm GL}(2^{k+1}, q)$, except when $k=0$ and $q\equiv 3\mod 4$. This follows from the description in Derek Holt's answer, but it can also be seen directly by observing that $T$ has the right order.
Write $N=S\times S$, so that $T= C_2 N$, and let $\chi\in {\rm Irr} (T)$. Three cases have to be considered:

1. We have $\chi_N \in {\rm Irr} (N)$. By induction, $\chi_N$ is afforded by a representation over its character value field. Since $N$ has a complement in $T$, this representation can be extended to a representation of $T$ over the same field.
2. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are not Galois conjugate. Then $\chi = \theta^T$ and $\theta$ have the same field of values and the same Schur index.
3. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are Galois conjugate. This means that $\chi=\theta^T$, but $|\mathbb{Q}(\theta):\mathbb{Q}(\chi)|=2$. Again, a representation $R\colon N\to {\rm GL}(\chi(1), \mathbb{Q}(\chi) )$ affording the character $\theta+\theta^g$ can be extended to a representation over the same field, since $N$ has a complement (of order 2) in $T$.
   To get the induction going when $q\equiv 3\mod 4$, one has to check that the Sylow 2-subgroup of ${\rm GL}(2,q)$ has trivial Schur indices, but that is clear since its a semidihedral group.

The irreducibles of the Sylow 2-subgroups of ${\rm GL}(n,q)$, $q$ odd, have indeed trivial Schur indices: Let $S$ be such a Sylow 2-subgroup. First, observe that the natural module $(\mathbb{F}_q)^n$ splits into a sum of simple modules $U_1\oplus \dotsb \oplus U_l$, and the dimension of each simple module is a power of $2$. This shows that $S \cong S_1 \times \dotsb \times S_l $, where the $S_i$'s are Sylow 2-subgroups of a ${\rm GL}(2^k, q)$. Thus, w.l.o.g. we may assume that $n=2^k$. Then I use induction on $k$. First note that if $S$ is a Sylow 2-subgroup of ${\rm GL}(2^k, q)$, then $$ T=\lbrace \begin{pmatrix} s & \\ & t \end{pmatrix} \mid s, t\in S \rbrace \cup \lbrace \begin{pmatrix} & s \\ t & \end{pmatrix} \mid s,t\in S\rbrace \cong S\wr C_2 $$ is a Sylow 2-subgroup of ${\rm GL}(2^{k+1}, q)$, except when $k=0$ and $q\equiv 3\mod 4$. This follows from the description in Derek Holt's answer, but it can also be seen directly by observing that $T$ has the right order.
Write $N=S\times S$, so that $T= C_2 N$, and let $\chi\in {\rm Irr} (T)$. Three cases have to be considered:

1. We have $\chi_N \in {\rm Irr} (N)$. By induction, $\chi_N$ has trivial Schur index. By Lemma~10.4 in Isaacs' character theory book, for example, it follows that $\chi$ has trivial Schur index.
2. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are not Galois conjugate. Then $\chi = \theta^T$ and $\theta$ have the same field of values and the same Schur index (again, see Lemma~10.4 in Isaacs).
3. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are Galois conjugate. This means that $\chi=\theta^T$, but $|\mathbb{Q}(\theta):\mathbb{Q}(\chi)|=2$. Now a representation $R\colon N\to {\rm GL}(\chi(1), \mathbb{Q}(\chi) )$ affording the character $\theta+\theta^g$ can be extended to a representation over the same field, since $N$ has a complement (of order 2) in $T$.
   To get the induction going when $q\equiv 3\mod 4$, one has to check that the Sylow 2-subgroup of ${\rm GL}(2,q)$ has trivial Schur indices, but that is clear since it's a semidihedral group.