The answer is 24.
Let $p$ be an odd prime. Using the binomial theorem, we have
$$(2 + \sqrt{5})^p + (2 - \sqrt{5})^p = \sum_{k=0}^{\frac{p-1}{2}} \binom{p}{2k} 2^{p+1-2k} \cdot 5^k = 2^{p+1} + \sum_{k=1}^{\frac{p-1}{2}} \binom{p}{2k} 2^{p+1-2k} \cdot 5^k,$$
which is an integer. Since $p$ is an odd prime, we have $(2 - \sqrt{5})^p < 0$. From this it follows that $|(2 + \sqrt{5})^p| = (2 + \sqrt{5})^p + (2 - \sqrt{5})^p$. Thus,
$$|(2 + \sqrt{5})^p| - 2^{p+1} = \sum_{k=1}^{\frac{p-1}{2}} \binom{p}{2k} 2^{p+1-2k} \cdot 5^k.$$
Since $\binom{p}{r}$ is divisible by $p$ for $1 \le r \le p-1$, $|(2 + \sqrt{5})^p| - 2^{p+1}$ is divisible by $p$.
It is easy to check that the expression is not divisible by $p$ when $p = 2$. Since there are 25 primes less than 100, the answer is 24.