Existence of $k$-splitting by Hilbert's Theorem 90? Or useless nonsense? I may have made by first successful cohomological argument.  Or not.  I don't know.  
Let $G$ be a connected, reductive quasisplit group over a (let's say perfect, although this probably doesn't matter) field $k$.  Let $\Gamma = \textrm{Gal}(\overline{k}/k)$.  Let $B = TU$ be a Borel subgroup with maximal torus $T$, both defined over $k$.  Let $\Delta$ be the base of $\Phi = \Phi(G,T)$ corresponding to $B$.  Choose a finite Galois extension $K$ of $k$ over which $T$ splits.
Choose a splitting, i.e. choice of isomorphisms $x_{\alpha}:\mathbf G_a \rightarrow U_{\alpha} : \alpha \in \Delta$ defined over $\overline{k}$.  On account of the fact that $\textrm{Ker Ad} = Z(G) = \bigcap\limits_{\alpha \in \Delta} \textrm{Ker } \alpha$, and the surjectivity of the mapping 
$$T(\overline{k}) \rightarrow \prod\limits_{\alpha \in \Delta} \overline{k}^{\ast}, t \mapsto (\alpha(t))_{\alpha}$$
it follows that any two splittings differ by $\textrm{Int } t$ for some $t \in T(\overline{k})$, and $t$ is unique modulo $Z(G)(\overline{k})$.
For $\sigma \in \Gamma_K = \textrm{Gal}(K/k)$, $\sigma.X_{\alpha} = \sigma \circ X_{\alpha} \circ \sigma^{-1}$ is an isomorphism $\mathbf G_a \rightarrow U_{\sigma.\alpha}$, so we can find a $t_{\sigma} \in T(\overline{k})$ (unique modulo $Z(G)(\overline{k})$) such that $\textrm{Int } t_{\sigma} \circ X_{\sigma.\alpha} = \sigma.X_{\alpha}$ for all $\alpha \in \Delta$.  We say that our splitting is defined over $k$ if actually $\sigma.X_{\alpha} = X_{\sigma.\alpha}$ for all $\sigma \in \Gamma_K$ and all $\alpha \in \Delta$.
My goal is to show that there exists a splitting which is defined over $k$.
With our initial choice of splitting, check that
$$t_{\sigma \tau}Z(G)(\overline{k}) = t_{\sigma} \sigma(t_{\tau})Z(G)(\overline{k})$$
Thus our initial choice of splitting $X_{\alpha}$ gives a $1$-cocycle $\Gamma)K \rightarrow T/Z(G)(\overline{k})$.
Check also that if we replace the $1$-cocycle above with an equivalent one, that amounts to just starting with a different splitting $X_{\alpha}'$, and conversely.
To say that a splitting is defined over $k$ is equivalent to the corresponding $1$-cocycle being trivial.
Now if $S$ is any torus (such as $T/Z(G)$) which splits over $K \supseteq k$, we have 
$$H^1(\Gamma_K, S(\overline{k})) = H^1(\Gamma,S(\overline{k})) = H^1(\Gamma, \varinjlim\limits_{E \supseteq K} S(E)) = \varinjlim\limits_{E \supseteq K} H^1(\Gamma, S(E))$$
where $E$ runs over all finite Galois extensions of $k$ containing $K$.  But for all such $E$, we have $S(E)$ isomorphic to a finite product of $E^{\ast}$s, and 
$$H^1(\Gamma, E^{\ast} \times \cdots \times E^{\ast}) = \prod\limits H^1(\Gamma, E^{\ast})$$
which is trivial by Hilbert's theorem 90.  This says that all $1$-cocycles are equivalent, i.e. there exists a splitting which is defined over $k$.  
 A: $\newcommand{\mf}[1]{\mathfrak{#1}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\Aut}{\mathrm{Aut}}$$\newcommand{\cont}{\mathrm{cont.}}$This is correct, I believe. I also wrote an argument for this once that looks a lot like yours. I include it below.
Let $B$ be a Borel subgroup of $G$ and $T$ a maximal torus of $B$. Note then that $(B_{\ov{F}},T_{\ov{F}})$ is certainly a $\Gamma_F$-stable Borel pair of $G_{\ov{F}}$. We then claim that there exists a $\Gamma_F$-stable pinning of $(B_{\ov{F}},T_{\ov{F}})$. To see this, let us decompose $\Delta^\ast(G_{\ov{F}},T_{\ov{F}},B_{\ov{F}})$ into $\Gamma_F$-orbits $\Delta_i$. For each $i$ choose an element $\alpha_i$ in $\Delta_i$. Let $\Gamma_{E_i}\subseteq\Gamma_F$ be the isotropy subgroup for $\alpha_i$ and fix representatives $\sigma_{i,1},\ldots,\sigma_{i,m_i}$ for $\Gamma_F/\Gamma_{E_i}$.
Let us note that since $\mf{g}_{\alpha_i}$ is stable under the action of $\Gamma_{E_i}$ on $\mf{g}\otimes_F\ov{F}$ that it contains a $\Gamma_{E_i}$-stable element. Indeed, fix $x\in\mf{g}_{\alpha_i}$ to be arbitrary. Then, for all $\sigma$ in $\Gamma_{E_i}$ one has that $\sigma(x)=c_\sigma x$ for some $c_\sigma\in\ov{F}$. Note then that $\sigma\mapsto c_\sigma$ gives an element of $Z^1_\cont(\Gamma_{E_i},\ov{F}^\times)$ which, by Hilbert's theorem 90, must be a $1$-coboundary. So, there exists some $c_0$ in $\ov{F}^\times$ such that $c_\sigma=c_0\sigma(c_0)^{-1}$. It's then easy to check that $c_0 x$ is a fixed element of $\mf{g}_{\alpha_i}$.
So, let us fix for each $i$ a $\Gamma_{E_i}$-fixed element $x_{\alpha_i}$ of $\mf{g}_{\alpha_i}$. Then, we define $x_{\sigma_{i,j}(\alpha_i)}:=\sigma_{i,j}(x_{\alpha_i})$. It's not hard to see that $\{x_{\sigma_{i,j}(\alpha_i)}\}$ is $\Gamma_F$-stable. Thus, $(B_{\ov{F}},T_{\ov{F}},\{x_{\sigma_{ij}(\alpha_i)}\}$ is $\Gamma_F$-stable splitting of $G_{\ov{F}}$ which, in particular, means that $\Aut(G_{\ov{F}},B_{\ov{F}},T_{\ov{F}},\{x_{\sigma_{ij}(\alpha_i)}\})$ is a $\Gamma_F$-stable subgroup of $\Aut(G_{\ov{F}})$.
