OLS does not require Gaussian features

A colleague rejects linear regression because a predictor is strongly non-Gaussian. Is that sufficient reason to reject the model?

No. OLS does not require predictors or the target to be normally distributed. Conditions on the conditional mean and errors determine interpretation and inference. Check functional form, exogeneity, leverage, dependence and heteroscedasticity. Gaussian errors are useful for exact finite-sample classical inference, not a requirement for computing a least-squares fit. Choose transformations for modelling reasons, not to force every marginal distribution into a bell curve.