**Yes, that is true.**
This is a consequence of the classical <A HREF="https://en.wikipedia.org/wiki/Trace_operator"><FONT FACE="Arial">Trace Theorem</FONT></A><FONT FACE="Arial">
 for Sobolev spaces. The proof can be found in any textbook on Sobolev spaces for example in Evans' <A HREF="https://mathscinet.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=RT&pg7=JOUR&pg8=ET&review_format=html&s4=evans&s5=partial&s6=&s7=&s8=Books&sort=Newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=2597943"><FONT FACE="Arial">Partial Differential Equations</FONT></A><FONT FACE="Arial">. The usual statement of the trace theorem deals with traces of 
$T:W^{1,p}(\Omega)\to L^p(\partial\Omega)$,
but the proof would also give traces 
$T:W^{1,p}([0,T]\times Y)\to L^p(\{ t\}\times Y)$. If $u\in W^{2,p}$, then the derivative is in $W^{1,p}$ so the trace of the derivative is in $L^p$ and since the derivative of the trace is the trace of the derivative (roughly speaking) we 
obtain the trace $T:W^{2,p}([0,T]\times Y)\to W^{1,p}(\{ t\}\times Y)$.