In order for H(z) to be a linear phase filter, it must zeros both on the inside of the unit circle and at the complementary locations (1/z) which are outside the unit circle. Therefore a linear phase circle has no stable causal inverse (since this would necessitate poles outside of the unit circle.)
Note that a linear phase filter can be decomposed into the cascade of a minimum phase filter (all zeros inside the unit circle) and a maximum phase filter with the same magnitude response (all zeros outside the unit circle). For this reason, only causal minimum phase FIR filters have a stable causal inverse.
Further interesting detail, the reverse filter (all the coefficients in reverse order) of a minimum phase filter IS the maximum phase filter. This is all now intuitive: the coefficients for a minimum phase filter will be dominant toward the start of the filter (resulting in the minimum delay as the signal will emerge from the filter sooner). Reversing this results in the same magnitude response but the filter will have the maximum delay as the signal will emerge from the filter later. When you cascade two filters, you convolve their coefficients, so we can also see how the cascade of a linear phase filter with a maximum phase filter results in symmetric coefficients. And it is well understood that any FIR filter with symmetric coefficients is a linear phase filter!
Example [2 1 1] is a minimum phase filter, and [1 1 2] is the reverse maximum phase filter. The cascade of the two results in the convolution of [2 1 1] and [1 1 2] which is [2 3 6 3 2] which is a linear phase filter.
If the filter has real coefficients, the all zeros must also have a complex conjugate. (But whether the filter has real or complex coefficients does not change the answer--- both real and complex linear phase filters will not have a stable causal inverse, nor will they have a stable anti-causal inverse).
Regarding causality and the ROC, for causal FIR filter the ROC is $|z|>0$ (all the poles are at $z = 0$), and for anti-causal filters it is $|z| < \infty$ (and all the poles are at $z = \infty$). In either case the requirement for zeros to be at reciprocal locations holds in order for the result to have linear phase.