Peggy,
Yes, in rereading your post, I did see you said you'd planted it in '99.
Our plants took maybe 3 years to decide to be the windbreak they were intended to be (but we knew we'd planted it in the right place when the wind kept knocking it down, LOL!) Then, again, we have several of them so after a certain point, the old canes all help to support each other. Is yours in an incredibly windy spot?
Does it have a lot of canes or are some of them actually dying to the ground?
If you want it to be upright, I don't understand why you'd cut out the old canes - they are what give the bush overall its structure and wind-resistance (if wind is part of the problem, as it definitely was here). Each year's new canes link with and get support from the old sturdy ones. Unless the old canes are actually dead, I would just leave them until the shrub overall develops enough strong, upright old wood to support itself...sounds like it isn't there yet. Also, Lois Hole's book says it blooms on old wood, so another reason to leave the old canes. We got an unusual amount of rain here too, but our roses aren't flopping. Weak,lax growth can be promoted by heavy fertilizing - something these hardy roses don't need.
Another thought...is it definitely a 'Therese Bugnet'? Pots often get mislabelled at nurseries. The lax habit over the long period you mentioned might suggest one of the "climbing" hardy roses, e.g. 'William Baffin', 'John Cabot', etc.. It might be worth googling some of these climbers...and then consider installing a trellis behind it, LOL!
A cursory description of 'Therese Bugnet', if it's helpful...
The bark on new growth is red and essentially thornless; despite the rugosa blood, the leaves are smooth; old stems have a thin covering of very narrow, needle-like thorns (nowhere near as thorny as many rugosa-types); very heavy bloom in spring, then sporadically until killing frost; flowers about 3" across...whew, that's enough!
I (or probably most of the other poeple on here, LOL!) could post pictures if there is any question of ID.
Q