The original version of this answer was backed up by facile intuition untempered by computation. I've now done the computation, and hopefully managed to come up with a correct formula that shows that we do have extensibility.
Linus correctly points out that this is a special case of the Witt extension theorem, which is the right answer, and I'm embarrassed not to have remembered that lovely result. As penance, I'll update my answer slightly to discuss the characteristic-$2$ case, and hope that I don't make any new mistakes along the way.
Write $b$ for the polar $(x, y) \mapsto q(x + y) - q(x) - q(y)$ of $q$, so that $2q(x)$ equals $b(x)$ for all $x \in V$. The superscript $\perp$ will mean orthogonal complements with respect to $b$.
$\DeclareMathOperator\Span{Span}$If $k$ is a field of characteristic $2$ and $V$ has odd dimension, then the radical of $q$ is a non-$q$-isotropic line $\ell$, and is the unique line preserved by all automorphisms of $q$. There is a different line $\ell'$ such that $H \mathrel{:=} \ell^{\prime\,\perp}$ is a hyperplane containing $\ell$ and $\ell'$, and $q\rvert_H$ admits an automorphism swapping them, which is therefore not extensible to an automorphism of $V$. (If $k$ is perfect, then we may take $\ell'$ to be any non-$q$-isotropic line.) For a concrete example, put $V = k^3$ and consider $q(x) = x_1 x_3 + x_2^2$ for all $x \in V$, in which case, with hopefully obvious notation, $\ell$ equals $\Span e_2$ and we may take, for example, $\ell' = \Span (e_1 + e_3)$, with the linear automorphism $x \mapsto x_1(e_2 + e_3) + x_2(e_1 + e_3) + x_3 e_3$ of $V$ restricting to an automorphism of $q\rvert_H$.
Thus the best we can do is to suppose that we are working over an arbitrary field $k$, and that $V$ is even-dimensional or $k$ does not have characteristic $2$. In either case, $b$ is non-degenerate.
As you have pointed out, we have extensibility when $b$ restricts to a non-degenerate pairing on $H$, so we may assume that it is degenerate. I suspect, though I haven't checked, that a similar argument works in characteristic $\ne 2$ and arbitrary codimension, as long as $H$ contains $H^\perp$; and, though again I haven't checked, that we can reduce to that case by first extending from $H$ to $H + H^\perp$.
Fix elements $v_0$ and $h^\perp$ of $V \setminus H$ and $H^\perp \setminus \{0\}$, and put $H_0 = \{v_0, h^\perp\}^\perp$. We may, and do, assume, after scaling $v_0$ and then replacing by a translate by an element of $H^\perp$ if necessary, that $q(v_0)$ equals $0$ and $b(v_0, h^\perp)$ equals $1$.
There are some $\lambda \in k^\times$ such that $\sigma(h^\perp)$ equals $\lambda^{-1}h^\perp$, and some $h_0 \in H_0$ such that $b(h_0, h)$ equals $b(v_0, \sigma(h))$ for all $h \in H_0$. Then, if I've managed to get the computation right this time, the linear extension of $\sigma$ to $V$ that sends $v_0$ to $\lambda (v_0 - \sigma(h_0) + q(h_0)h^\perp)$ is an automorphism of $q$.
