A Kolmogorov-Smirnov test will do the job well.
This procedure compares data to a prespecified distribution: in this case, the uniform distribution on the interval $[0, 90].$ Its test statistic often called "$D$") is the largest vertical deviation between the empirical cumulative distribution function (ecdf) of the data and the cumulative distribution function (cdf) of the reference distribution.
To assess how well this might work, I generated Normal random variates with mean $25$ and standard deviation $28,$ reduced them modulo $180$ degrees to lie between $-90$ and $90$ (thereby simulating the observed orientation differences), and took their absolute values (to simulate the observed absolute orientation differences). In the plots below, the upper left plot graphs the density function and the bottom left plot is an example of a histogram of a sample of size $1350$ -- roughly the same size as displayed in the question.
At the bottom right are the ecdf of this sample (black) and the reference cdf (red). An idealization of this plot appears in the upper right, which compares the two cdfs. That plot includes a dashed vertical line marking the location of their greatest vertical departure.
This is such a large sample that the K-S test is virtually guaranteed to discriminate between the two distributions I have used. As an example of what could happen in a more delicate situation, here are the results for a sample of just $50$ observations where the standard deviation of the underlying distribution is 1.5 times greater ($42$ instead of $28$).
In 10,000 simulations of this situation, the K-S test found a significant difference (at the usual 95% threshold) only about half the time (54% of them). This estimates the power of the test to detect the difference in this situation.
These tests were carried out using the
ks.test function, as in this line and its output. Its input is the array
X of data values shown in histogram of the preceding figure:
ks.test(X, punif, min=0, max=90)
One-sample Kolmogorov-Smirnov test
D = 0.23227, p-value = 0.007441
alternative hypothesis: two-sided