Can I expect 18O-hydroxide when adding slight amount of NaOH-16O to 18O-water?
Yes, it will but it will be a hydrogen exchange, not an oxygen exchange.
$$\ce{2 H2^{18}O &lt;=&gt; \color{blue}{H3^{18}O+} + ^{18}OH-}$$
$$\ce{\color{blue}{H3^{18}O+} + ^{16}OH- &lt;=&gt; H2^{18}O + H2^{16}O }$$
The oxygen will mix with water and hydroxide to a near equivalent ratio of $\ce{^{18}O:^{16}O}$ in both $\ce{H2O}$ and $\ce{OH-}$ but not perfectly equivalent as isotopes do not have equivalent chemical potentials.