I think here is one example. For each rational in $[0, 1]$ of the form $k/2^{n}$, let $S_{k, n} := \{k/2^n\} \times  \{-2^{-n}, 2^{-n}\}$, and define $S = \cup_{k \in \mathbb Z, n \in \mathbb Z_+}\ \ S_{k, n} \cup ([0, 1] \times \{0\})$.

 Identifying the latter set in the union  can be shown that the set of $x \in [0, 1]$ with the liminf in question $0$ for all $v \in \mathbb R^2$ is the set $E$ defined as follows:

Denote by $L_k (x)$ the length of the string of $0$’s or $1$’s beginning at the $k$’th decimal place of the binary expansion of $x$. Set 

$$E := \{x \in [0, 1]| \ \limsup_{k \to \infty} L_k (x) - k = +\infty\}.$$

This set $E$ has full measure in $[0, 1]$, whence the set of $x \in S$. with $T_S (x) = \mathbb R^2$ is of $\mathcal H^1$ measure $1$. 

Finally we note that $S$ is closed and has $\mathcal H^1$ measure $1$, and so $S$ satisfies the requirements of your problem.

Edit: Ah, the construction as stated does not work, but I believe something similar might. I’ll leave it up since I will try again to construct it later tonight, possibly with a calculation to show it works.