As pointed out by @whuber, the expression for the Copula is
$$C(u,v) = \big[\max \{u^{-\theta}+v^{-\theta}-1,\;0\}\big]^{-1/\theta} , \;\;\theta \in [-1,\infty), \theta\neq 0 $$
When $\theta >0 \Rightarrow u^{-\theta}+v^{-\theta}-1>0$ for the whole joint support, and we can "ignore" the $\max$ operator. But when $\theta<0$ the $\max$ operator becomes effective, and this must be taken into account when performing the integration, since for subsets of the joint support the value of the copula is zero.
An alternative approach could be the following:
Treat the Copula itself as a (univariate) random variable (which in turn is a function of two other random variables):
$$T = C(U,V)$$
Then $T$ has a distribution function called "Kendall distribution function", and it is equal to
$$K_C(t) = t- \frac {\varphi (t)}{\varphi'(t)} \tag{1}$$
where $\varphi(t)$ is the copula's generator function, and the prime denotes the first derivative.
The relation between Kendall's tau and a copula is
$$\tau = 4 \cdot E[C(U,V)] - 1 \tag{2}$$
and from $(1)$ we have
$$E[C(U,V)] = E(T) = \int_0^1tdK_C(t)$$
Integrating by parts we have
$$\int_0^1tdK_C(t) = tK_C(t)\Big|_0^1 - \int_0^1K_C(t)dt = 1 -\int_0^1K_C(t)dt$$
So Kendall's tau now is
$$\tau = 4\cdot \left(1 -\int_0^1K_C(t)dt\right) - 1 = 3 - 4\int_0^1K_C(t)dt$$
Inserting the expression for $K_C(t)$ we have
$$\tau = 3 - 4\int_0^1\left[t- \frac {\varphi (t)}{\varphi'(t)}\right]dt = 1 + 4\int_0^1\frac {\varphi (t)}{\varphi'(t)}dt$$
which is a general expression for the calculation of Kendall's tau related to a copula.
For Clayton's Copula, its generator function is (for $\theta \neq 0$)
$$\varphi(t) = \frac1{\theta}\left(t^{-\theta}-1\right)$$
Completing the calculations, one arrives at $\tau = \theta / (\theta + 2)$.
Then , for $\theta = -1/2$ we have indeed that $\tau = -1/3$.
Best Answer
This paper discusses the contexts where you can and can't use a normal approximation for Tau. According to Wikipedia, it also looks like the validity of normal/Z depends on how your version of Tau handles ties. My sense is that it's probably safer not to assume that it's Gaussian, especially with relatively low sample sizes.
I couldn't think of a reason why Kendall's Tau wouldn't be compatible with the bootstrap, but I wasn't 100% sure. So I looked it up:
Here's a paper by Brad Efron, the inventor of the bootstrap, that uses it for Tau (Section 5).
Here's a paper that spends some time discussing the bootstrap in the context of Tau (mostly Section 4).
Looks like you shouldn't have any serious problems using the bootstrap for tau.