In fact, $O(n^2)$ arithmetic operations is not avoidable, in general. A general result by Windograd Winograd (see "Algebraic complexity theory" by Buergisser, Clausen, and Shokrollahi, Sect. 13.2) shows that the for a generic $m\times n$ matrix (all the entries are algebraically independent) and a generic vector one will need $O(mn)$ multiplications and additions (there is a precise formula too, but $O(mn)$ would do for our purposes). PSD matrices are symmetric, so directly it won't apply t $H$. However, you can cut $H$ into blocks: $H=\begin{pmatrix} H_{11}&H_{12}\\ H_{12}^\top&H_{22}\end{pmatrix}$. Then $H_{12}$ is generic, in general, and, writing $v$ as a block vector $v=(v^1,v^2)$ one has $b=(b^1,b^2)=(H_{11}v^1+H_{12}v^2, H_{12}^\top v^1+H_{22}v^2)$. Now Winograd's result is applicable to $H_{12}$, so you still get $O(n^2)$, as you cannot avoid computing $H_{12}v^2$ when you try to get $b$.
In fact, $O(n^2)$ arithmetic operations is not avoidable, in general. A general result by Windograd (see "Algebraic complexity theory" by Buergisser, Clausen, and Shokrollahi, Sect. 13.2) shows that the for a generic $m\times n$ matrix (all the entries are algebraically independent) and a generic vector one will need $O(mn)$ multiplications and additions (there is a precise formula too, but $O(mn)$ would do for our purposes). PSD matrices are symmetric, so directly it won't apply t $H$. However, you can cut $H$ into blocks: $H=\begin{pmatrix} H_{11}&H_{12}\\ H_{12}^\top&H_{22}\end{pmatrix}$. Then $H_{12}$ is generic, in general, and, writing $v$ as a block vector $v=(v^1,v^2)$ one has $b=(b^1,b^2)=(H_{11}v^1+H_{12}v^2, H_{12}^\top v^1+H_{22}v^2)$. Now Winograd's result is applicable to $H_{12}$, so you still get $O(n^2)$, as you cannot avoid computing $H_{12}v^2$ when you try to get $b$.