As mentioned by Tendero in a comment, you should be interested in the stability of the closed-loop transfer function
where I've assumed unity feedback. Since the stability of $(1)$ is determined by the zeros of the denominator $1+H(s)$, we're interested in the encirclement of the point $-1+j0$ (and not in the encirclement of the origin). In your example, the number of clockwise encirclements of the point $-1+j0$ happens to equal the number of encirclements of the origin, i.e., we have $N=1$. Since the given $H(s)$ has no poles in the right half-plane, the number of zeros in the right half-plane equals the number of clockwise encirclements, i.e., $Z=N=1$, which means that $Q(s)$ has one pole in the right half-plane, and, consequently, the closed-loop transfer function is unstable.
This is also easily verified by computing the poles of $Q(s)$, which are the roots of the polynomial $s^4+s^3-1$:
-1.38028 + 0.00000i
-0.21945 + 0.91447i
-0.21945 - 0.91447i
0.81917 + 0.00000i
This shows that one of the four roots is in the right half-plane, as predicted by the Nyquist plot.