Overall, the answer to the question in the title is: It depends. It depends on the distribution of $\hat \beta$ and the level of significance $\alpha$. But the general sense you are observing holds.
Now, to be specific, suppose $\frac{\hat \beta - \beta_0}{\text{SE}(\hat \beta)}$ has a central $t$ distribution (which is symmetric about 0). Then if $\hat \beta - \beta_0$ is negative in the case of the alternative hypothesis $\beta > \beta_0$, or if $\hat \beta - \beta_0$ is positive in the case of the alternative hypothesis $\beta < \beta_0$, the $p$-value will be at least 0.5. Unless we are willing to tolerate a large type-I error rate (i.e., an $\alpha$ even bigger than the $p$-value which already is higher than 0.5), yes, we will fail to reject $H_0$.
Philosophy of hypothesis testing gives more credence to $H_0$ in the sense that unless strong evidence (completely clearing "reasonable doubt") is presented by the data, we will not question $H_0$ which might be the current standard. This can be seen also from P(Type-I Error) being controlled first.
One last note: One cannot truly compare a 2-sided confidence interval to a 1-tailed hypothesis testing decision. But, to clear your intuitive thinking, if confidence interval overlaps then the hypothesis testing decision theory essentially says we cant be sure - so let's just not jump the gun (i.e., stick with $H_0$, our current belief state). Confidence interval will have to "completely clear the $H_0$ area" (loosely speaking) for us to feel strongly about the evidence provided by the data in the direction of the alternative hypothesis. Again, keep in mind, 2-sided CI cannot be exactly compared to 1-tailed test.
