Yes, there are some simple relationships between confidence interval comparisons and hypothesis tests in a wide range of practical settings. However, in addition to verifying the CI procedures and t-test are appropriate for our data, we must check that the sample sizes are not too different and that the two sets have similar standard deviations. We also should not attempt to derive highly precise p-values from comparing two confidence intervals, but should be glad to develop effective approximations.
In trying to reconcile the two replies already given (by @John and @Brett), it helps to be mathematically explicit. A formula for a symmetric two-sided confidence interval appropriate for the setting of this question is
$$\text{CI} = m \pm \frac{t_\alpha(n) s}{\sqrt{n}}$$
where $m$ is the sample mean of $n$ independent observations, $s$ is the sample standard deviation, $2\alpha$ is the desired test size (maximum false positive rate), and $t_\alpha(n)$ is the upper $1-\alpha$ percentile of the Student t distribution with $n-1$ degrees of freedom. (This slight deviation from conventional notation simplifies the exposition by obviating any need to fuss over the $n$ vs $n-1$ distinction, which will be inconsequential anyway.)
Using subscripts $1$ and $2$ to distinguish two independent sets of data for comparison, with $1$ corresponding to the larger of the two means, a non-overlap of confidence intervals is expressed by the inequality (lower confidence limit 1) $\gt$ (upper confidence limit 2); viz.,
$$m_1 - \frac{t_\alpha(n_1) s_1}{\sqrt{n_1}} \gt m_2 + \frac{t_\alpha(n_2) s_2}{\sqrt{n_2}}.$$
This can be made to look like the t-statistic of the corresponding hypothesis test (to compare the two means) with simple algebraic manipulations, yielding
$$\frac{m_1-m_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \gt \frac{s_1\sqrt{n_2}t_\alpha(n_1) + s_2\sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 s_2^2 + n_2 s_1^2}}.$$
The left hand side is the statistic used in the hypothesis test; it is usually compared to a percentile of a Student t distribution with $n_1+n_2$ degrees of freedom: that is, to $t_\alpha(n_1+n_2)$. The right hand side is a biased weighted average of the original t distribution percentiles.
The analysis so far justifies the reply by @Brett: there appears to be no simple relationship available. However, let's probe further. I am inspired to do so because, intuitively, a non-overlap of confidence intervals ought to say something!
First, notice that this form of the hypothesis test is valid only when we expect $s_1$ and $s_2$ to be at least approximately equal. (Otherwise we face the notorious Behrens-Fisher problem and its complexities.) Upon checking the approximate equality of the $s_i$, we could then create an approximate simplification in the form
$$\frac{m_1-m_2}{s\sqrt{1/n_1 + 1/n_2}} \gt \frac{\sqrt{n_2}t_\alpha(n_1) + \sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 + n_2}}.$$
Here, $s \approx s_1 \approx s_2$. Realistically, we should not expect this informal comparison of confidence limits to have the same size as $\alpha$. Our question then is whether there exists an $\alpha'$ such that the right hand side is (at least approximately) equal to the correct t statistic. Namely, for what $\alpha'$ is it the case that
$$t_{\alpha'}(n_1+n_2) = \frac{\sqrt{n_2}t_\alpha(n_1) + \sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 + n_2}}\text{?}$$
It turns out that for equal sample sizes, $\alpha$ and $\alpha'$ are connected (to pretty high accuracy) by a power law. For instance, here is a log-log plot of the two for the cases $n_1=n_2=2$ (lowest blue line), $n_1=n_2=5$ (middle red line), $n_1=n_2=\infty$ (highest gold line). The middle green dashed line is an approximation described below. The straightness of these curves belies a power law. It varies with $n=n_1=n_2$, but not much.
The answer does depend on the set $\{n_1, n_2\}$, but it is natural to wonder how much it really varies with changes in the sample sizes. In particular, we could hope that for moderate to large sample sizes (maybe $n_1 \ge 10, n_2 \ge 10$ or thereabouts) the sample size makes little difference. In this case, we could develop a quantitative way to relate $\alpha'$ to $\alpha$.
This approach turns out to work provided the sample sizes are not too different from each other. In the spirit of simplicity, I will report an omnibus formula for computing the test size $\alpha'$ corresponding to the confidence interval size $\alpha$. It is
$$\alpha' \approx e \alpha^{1.91};$$
that is,
$$\alpha' \approx \exp(1 + 1.91\log(\alpha)).$$
This formula works reasonably well in these common situations:
Both sample sizes are close to each other, $n_1 \approx n_2$, and $\alpha$ is not too extreme ($\alpha \gt .001$ or so).
One sample size is within about three times the other and the smallest isn't too small (roughly, greater than $10$) and again $\alpha$ is not too extreme.
One sample size is within three times the other and $\alpha \gt .02$ or so.
The relative error (correct value divided by the approximation) in the first situation is plotted here, with the lower (blue) line showing the case $n_1=n_2=2$, the middle (red) line the case $n_1=n_2=5$, and the upper (gold) line the case $n_1=n_2=\infty$. Interpolating between the latter two, we see that the approximation is excellent for a wide range of practical values of $\alpha$ when sample sizes are moderate (around 5-50) and otherwise is reasonably good.
This is more than good enough for eyeballing a bunch of confidence intervals.
To summarize, the failure of two $2\alpha$-size confidence intervals of means to overlap is significant evidence of a difference in means at a level equal to $2e \alpha^{1.91}$, provided the two samples have approximately equal standard deviations and are approximately the same size.
I'll end with a tabulation of the approximation for common values of $2\alpha$. In the left hand column is the nominal size $2\alpha$ of the original confidence interval; in the right hand column is the actual size $2\alpha^\prime$ of the comparison of two such intervals:
$$\begin{array}{ll}
2\alpha & 2\alpha^\prime \\ \hline
0.1 &0.02\\
0.05 &0.005\\
0.01 &0.0002\\
0.005 &0.00006\\
\end{array}$$
For example, when a pair of two-sided 95% CIs ($2\alpha=.05$) for samples of approximately equal sizes do not overlap, we should take the means to be significantly different, $p \lt .005$. The correct p-value (for equal sample sizes $n$) actually lies between $.0037$ ($n=2$) and $.0056$ ($n=\infty$).
This result justifies (and I hope improves upon) the reply by @John. Thus, although the previous replies appear to be in conflict, both are (in their own ways) correct.
@Placidia is right. However, this is a simplified case because you do not have any interactions. That is, your model assumes (rightly or wrongly) that moving from LandUseHigh to LandUseLow is associated with an increase of 0.25 units in your response variable no matter what the level of Type_LU is or what the values of your continuous covariates are.
If you did have interactions, then the coefficient on LandUseLow would correspond to the change from LandUseHigh only when Type_LU is set at the reference level. Likewise, if there were an interaction between LandUse and a continuous covariate, the coefficient on LandUseLow would correspond to the change from LandUseHigh only when the value of the continuous covariate were $0$. If you had a three-way interaction between LandUse, Type_LU and a continuous covariate, it would indicate the change when both Type_LU is set at the reference level and the value of the continuous covariate were $0$.
Update: The (intercept) indicates the value of the reference category. You have two categorical variables, so you have two reference categories. Your reference categories are LandUseHigh and an unspecified level of Type_LU (which I assume you know). So the value of the Estimate in the (intercept) row is the predicted mean for those study units in both of those categories when all the continuous covariates are equal to $0$. Again, because you don't have an interaction term, the value of LandUseHigh when Type_LU is conserva, private, or state is the sum of the estimate for the intercept plus the estimate for the appropriate level of Type_LU.
Best Answer
I'd like to refine and comment on some of your statements.
First, to start with the mechanics, depending on the sign of the difference between the estimate and the mean under the null hypothesis, the one-sided p-value is either the two-sided pvalue divided by 2, or the complement of that value. For your case, the null is that the mean is $0$, and your estimate of $d$ is positive, and so you indeed take the pvalue divided by two as you did. Mechanically, that is fine.
Now to answer
the answer is that you are mostly correct, but only conditional on satisfying assumptions required for your test. These assumptions include some statistical ones relating to using a t-test and your linear regression model, and I won't go into these because it's standard to assume them in most cases. However, another key assumption is that you did not choose to do a one-sided test conditional on seeing that the two-sided test is not significant at your chosen level. If you were to do that, then the p-value you get loses much meaning, and you would certainly not be able to conclude what you said about the estimate.
In general, it is quite unconventional to perform one-sided tests, and it is especially concerning when the two-sided test fails to reject the null at your given significance level, but the one-sided test does reject the null. If you were to report the one-sided test in a table, you would have to make it extremely clear that you are indeed performing a one-sided test, and I guarantee most scientific journals will question that decision, and be further critical when they realize that the two-sided test is not significant. Why are you using a one-sided test? Your question is about stock returns, and they can easily be negative. I would be extremely careful and wary about performing a one-sided test here...
EDIT:
To answer your comment, you can typically make the same conclusions rejecting the null under a two-sided test as you would with rejecting the null under a one-sided test. Under a two-sided test, if you reject the null, then you conclude the effect is significantly different from the null value, and the effect is in the direction of the estimate. So in your case, compared to $0$, a positive value being significant using a two-sided test would let you conclude exactly what you wanted.
Think of a one-sided test as 'buying information' and the cost is that you cannot detect any difference on the other side of what you posit with the one-sided test. Recall that you need to come up with your hypothesis before observing the data, so in your case, if had decided to do a one-sided test of the effect being positive and you observed a negative effect, you would not be able to say anything about it, because by starting off with a one-sided positive test, you already assumed that a negative value is impossible! And modifying the test after the fact to be one-sided negative (or even two-sided) is wrong, and you lose the ability to read into your p-value. Since it is very rare to truly know beforehand the sign of the estimate (intuition/experience is not good enough, because then you will just be confirming your biases without ever testing them), you should almost always avoid one-sided tests. But rejecting the null of a two-sided test corresponds to what you would expect: you reject the null, and the effect is in the direction that you observe (so in your case, greater than the null of $0$).