Remark. The following answer uses the same methodology as the accepted leading answer. It is self-contained, includes Maple code and some asymptotics. Perhaps it can serve a didactic purpose and show that the basic computation is very simple.
With $z$ for successes and $w$ for failures we get the generating function
$\left(1+z+\cdots+u\frac{z^m}{1-z}\right) \left(\sum_{q\ge 0} \frac{w^q}{(1-w)^q} \left(z+\cdots+u\frac{z^m}{1-z}\right)^q \right)\frac{1}{1-w}.$
Now as a sanity check when we put $w=z$ and $u=1$ we should get all binary strings. Doing the calculation we find
$\frac{1}{1-z} \frac{1}{1-wz/(1-w)/(1-z)} \frac{1}{1-w} \\ = \frac{1}{(1-w)(1-z)-wz} = \frac{1}{1-w-z}.$
Putting $z=w$ yields
$\frac{1}{(1-z)^2} \frac{1}{1-z^2/(1-z)^2} = \frac{1}{(1-z)^2-z^2} = \frac{1}{1-2z}$
and the sanity check goes through.
Now for the probabilities we must subtract the value for $u=0$ from the generating function and then set $u=1.$ We obtain for the zero term
$\frac{1-z^m}{1-z} \left(\sum_{q\ge 0} \frac{w^q}{(1-w)^q} z^q \frac{(1-z^{m-1})^q}{(1-z)^q}\right) \frac{1}{1-w} \\ = \frac{1-z^m}{1-z} \frac{1}{1-wz(1-z^{m-1})/(1-w)/(1-z)} \frac{1}{1-w} = \frac{1-z^m}{(1-w)(1-z)-wz(1-z^{m-1})} = \frac{1-z^m}{1-w-z+wz^{m}}.$
To obtain the probabilities set $z=pv$ and $w=(1-p)v$ to get for the one term
$\frac{1}{1-w-z} = \frac{1}{1-(1-p)v-pv} = \frac{1}{1-v}$
so that $[v^n] \frac{1}{1-v} = 1.$
The subtracted contribution from the zero term is
$\frac{1- p^m v^m}{1-v+(1-p)p^m v^{m+1}}.$
Now for coefficient extraction we write
$[v^Q] \frac{1}{1-v+(1-p)p^m v^{m+1}} \\ = [v^Q] \sum_{q\ge 0} (v-(1-p)p^m v^{m+1})^q \\ = [v^Q] \sum_{q\ge 0} \sum_{r=0}^q {q\choose r} v^{q-r} (-1)^r (1-p)^r p^{mr} v^{(m+1)r}. \\ = [v^Q] \sum_{q\ge 0} \sum_{r=0}^q {q\choose r} v^{mr+q} (-1)^r (1-p)^r p^{mr} \\ = \sum_{r=0}^{\lfloor Q/m\rfloor} {Q-mr\choose r} (-1)^r (1-p)^r p^{mr}.$
We thus get the closed form
$\bbox[5px,border:2px solid #00A000]{ 1 - a_N + p^m a_{N-m} \quad \text{where} \quad a_Q = \sum_{r=0}^{\lfloor Q/m\rfloor} {Q-mr\choose r} (-1)^r (1-p)^r p^{mr}.}$
Note that for $N\lt m$ we get
$1- {N\choose 0} \times 1 \times 1\times 1 = 0$
which is the correct result. Also for $N=m$ we obtain
$1 - 1 - {0\choose 1} \times -1 \times (1-p) \times p^m + p^m \times 1 = p^m $
which is correct as well.
Now for an asymptotic we note that the root $\rho$ with the smalltest modulus of $1-v+(1-p)p^m v^{m+1}$ is the one close to $v_0=1$ and we get from Newton-Raphson the first approximation
$v_1 = v_0 - \frac{1-v_0 + (1-p)p^m v_0^{m+1}}{-1+(m+1)(1-p)p^m v_0^{m}}.$
This works out to
$1+ \frac{(1-p)p^m}{1-(m+1)(1-p)p^m} = \frac{1-m(1-p)p^m}{1-(m+1)(1-p)p^m} = \rho.$
The corresponding term from the partial fraction decomposition is
$\frac{1}{v-\rho} \mathrm{Res}_{v=\rho} \frac{1}{1-v+(1-p)p^m v^{m+1}} \\ = \frac{1}{v-\rho} \times \left.\frac{1}{-1+(m+1)(1-p)p^m v^{m}}\right|_{v=\rho} \\ = - \frac{1}{\rho} \frac{1}{1-v/\rho} \frac{1}{-1+(m+1)(1-p)p^m \rho^{m}}.$
We thus obtain
$\bbox[5px,border:2px solid #00A000]{ a_Q \sim \frac{1}{\rho^{Q+1}} \frac{1}{1-(m+1)(1-p)p^m \rho^{m}}.}$
We may replace $\rho$ by a better approximation from Newton-Raphson in a setting where we require numerics, which is done in the sample code which now follows.
MRUNPROB := proc(N, m) option remember; local ind, d, pos, cur, run, runs, prob, zcnt, wcnt; prob := 0; for ind from 2^N to 2*2^N-1 do d := convert(ind, base, 2); cur := -1; pos := 1; run := []; runs := []; while pos <= N do if d[pos] <> cur then if nops(run) > 0 then runs := [op(runs), [run[1], nops(run)]]; fi; cur := d[pos]; run := [cur]; else run := [op(run), cur]; fi; pos := pos + 1; od; runs := [op(runs), [run[1], nops(run)]]; if nops(select(r -> (r[1] = 1 and r[2] >= m), runs)) > 0 then wcnt := add(`if`(r[1] = 0, r[2], 0), r in runs); zcnt := add(`if`(r[1] = 1, r[2], 0), r in runs); prob := prob + p^zcnt * (1-p)^wcnt; fi; od; expand(prob); end; V1 := proc(N, m) option remember; local gf; gf := (1-p^m*v^m)/(1-v+(1-p)*p^m*v^(m+1)); expand(1-coeftayl(gf, v=0, N)); end; a := (Q, m) -> add(binomial(Q-m*r,r)*(-1)^r*(1-p)^r*p^(m*r), r = 0 .. floor(Q/m)); V2 := proc(N, m) option remember; expand(1-a(N,m)+p^m*a(N-m,m)); end; a2 := proc(Q, m) local rho; rho := (1-m*(1-p)*p^m)/(1-(m+1)*(1-p)*p^m); 1/rho^(Q+1)*1/(1-(m+1)*(1-p)*p^m*rho^m); end; a3 := proc(Q, m, p) local rho; rho := sort([fsolve(1-v + (1-p)*p^m*v^(m+1), v)], (r1, r2) -> abs(r1-1) < abs(r2-1)); rho := op(1, rho); 1/rho^(Q+1)*1/(1-(m+1)*(1-p)*p^m*rho^m); end;