Does using Normal Order Evaluation instead of Normal Order Reduction lose the Normalization theorem? Normal Order Reduction (NOR)
reduce the leftmost, outermost redex.

Normal Order Evaluation (NOE)
reduce the leftmost, outermost redex, but not within the body of abstractions.

So (λw. (λx.x) z) is in normal form under NOE, but not under NOR.

Does using NOE instead of NOR weaken the Normalization property?

The Normalization property states that if there is a normal form under beta-reduction, that NOR will find it.
Edit:
Sorry, it's the Curry/Feys Theorem that says that NOR always finds a normal form if it exists.
NOE reduction strategy (as you define it) won't find the $\beta$-normal form of a term. For example, the normal form of $\lambda x.((\lambda y.y)x)$ is $\lambda x.x$
but NOE won't find it, because it won't reduce the inner redex $(\lambda y.y)x$ which is under the outer $\lambda$-abstraction. So it's not a normalizing reduction strategy.

However, in a typed functional language with data types like Int, if you know that some expression is of such a type, you know that it cannot be a function, so it's enough to restrict reductions to so-called Weak Head Normal Form (which is close to NOE), and it simplifies many things. See also:


http://www.haskell.org/haskellwiki/Weak_head_normal_form
https://en.wikibooks.org/wiki/Haskell/Graph_reduction#Weak_Head_Normal_Form
https://stackoverflow.com/questions/6872898/haskell-what-is-weak-head-normal-form