What is the difference in the reducing power of various ionic hydrides?
This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)

The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.

Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.

In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).

In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.

Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).

At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....

Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.

As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant. 

So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...

As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.

Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.