normal $\; \not\Rightarrow \;$ irrationality measure $2$
There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud  (2002). For related results, see  and .
irrationality measure $2$ $\; \not\Rightarrow \;$ normal
There exist numbers with irrationality measure $2$ that are not normal. In fact, there exist numbers with irrationality measure $2$ that fail to be simply normal in every base. Note that this is stronger than failing to be simply normal in base $10,$ which in turn is stronger than failing to be normal in base $10,$ which in turn is stronger than "normal" in the sense that you are asking about. Shallit  (1979) showed that the continued fraction expansion of
has bounded partial quotients (see Theorem 3 on p. 213 for meaning of $B(u,\infty)$ and Theorem 9 on p. 216 for the result), and thus has irrationality measure $2.$ [Indeed, for this we only need the $n$th partial quotient to be bounded by a linear function of $n.$ See Robert Israel's answer to the math overflow question Numbers with known irrationality measures.] However, it is clear that this number is not simply normal in any base. Indeed, in this number's expansion in any base, the proportion of $0$'s approaches $1,$ and hence the proportions for each of the other digits approaches $0.$
 Yann Bugeaud, Nombres de Liouville et nombres normaux [Liouville numbers and normal numbers], Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences 335 #2 (2002), 117-120.
 Verónica Becher, Pablo Heiber and Theodore A. Slaman, A computable absolutely normal Liouville number, preprint, 30 January 2014, 14 pages.
 Satyadev Nandakumar and Santhosh Kumar Vangapelli, Normality and finite-state dimension of Liouville numbers, arXiv:1204.4104v2, 21 January 2014.
 Jeffrey Outlaw Shallit, Simple continued fractions for some irrational numbers, Journal of Number Theory 11 #2 (April 1979), 209-217.