Suppose you have a beta prior with parameters $\alpha$ and $\beta$. You draw $n \geq 1$ points from a binomial distribution and observe $k > n/2$ successes. Does your posterior probability in the interval $[1/2,1]$ increase? Intuitively, I think ``yes'', but I am bad at evaluating integrals. Since the beta family is conjugate to the binomial distribution, I believe that we can rephrase the question as follows. If $k$ and $n$ are positive integers such that $1 \leq k \leq n$ and $k > (n-k)$, is the following true?
$$\frac{1}{B(\alpha, \beta)} \int_{1/2}^{1} x^{\alpha-1}(1-x)^{\beta-1}dx < \frac{1}{B(\alpha+k, \beta+(n-k))} \int_{1/2}^{1} x^{\alpha+k-1}(1-x)^{\beta+n-k-1}dx$$
where $B(\cdot, \cdot)$ is the beta-function.
More generally, for $r \in [1/2,1)$, if one observes $k$ many successes in $n$ trials and $k/n > r$, does it follow that one's posterior in the interval $[r,1]$ increases?
No - suppose that your prior made you confident that $p$ was large. For example, suppose that $\alpha = 10, \beta = 1$. Then, if you saw 5 successes out of 9 trials, you would actually be less confident that $p$ was so large, and your posterior probability that $p > 1/2$ would be less than your prior probability that $p > 1/2$. (These choices of $\alpha, \beta, k, n$ were just a guess I made that happened to work.)
Letting $\alpha = 10, \beta = 1, k = 5, n = 9$, we have $$\frac{1}{B(\alpha, \beta)} \int_{1/2}^{1} x^{\alpha-1}(1-x)^{\beta-1}dx = \frac{1}{B(10, 1)} \int_{1/2}^{1} x^{10-1}(1-x)^{1-1}dx = 0.9990234,$$ and $$\frac{1}{B(\alpha+k, \beta+(n-k))} \int_{1/2}^{1} x^{\alpha+k-1}(1-x)^{\beta+n-k-1}dx = \frac{1}{B(15, 5)} \int_{1/2}^{1} x^{15-1}(1-x)^{5-1}dx = 0.9903946.$$ So the left-hand side of your formula is greater than the right-hand side in this case.